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Maharashtra SSC Class 10 Mathematics Chapter 2 Quadratic Equations: textbook solutions

Solutions written by Shishya, checked, not the board's official answers

Step-by-step working for the practice-set questions of Chapter 2, “Quadratic Equations”, in Mathematics Part I (Algebra), Standard X (Maharashtra SSC, English medium; the chapter is on pages 33-54 of the printed book). Each question is in our own short words and cited by its set and number; open the textbook for the exact wording and figures.

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How these were checked: drafted with AI on 28 Oct 2026. A solution is shown only if a calculator check recomputed every final value from the question's own numbers and every arithmetic step in the working agreed. 73 of 76 questions in this chapter passed; the others are listed below as not solved here. These are not the board's official answers: the textbook and your teacher are the authority, and if something here disagrees with the book, follow the book and tell us.

Textbook PDF read on 28 Oct 2026 from the publisher's official site.

Practice Set 2.1

Not solved here yet: Q1 (see the textbook).

Practice Set 2.1, Q2
Which of these are quadratic equations? (1) x² + 5x − 2 = 0 (2) y² = 5y − 10 (3) y² + 1/y = 2 (4) x + 1/x = −2 (5) (m + 2)(m − 5) = 0 (6) m³ + 3m² − 2 = 3m³
1. A quadratic equation is a polynomial equation of degree 2 in one variable.
2. (1) The highest power of x is 2, so it is a quadratic equation.
3. (2) Rearranging gives y² − 5y + 10 = 0, which has degree 2, so it is a quadratic equation.
4. (3) Multiplying by y gives y³ + 1 = 2y, or y³ − 2y + 1 = 0; the highest power is 3, so it is not a quadratic equation.
5. (4) Multiplying by x gives x² + 1 = −2x, or x² + 2x + 1 = 0, which has degree 2, so it is a quadratic equation.
6. (5) Expanding gives m² − 3m − 10 = 0, which has degree 2, so it is a quadratic equation.
7. (6) Moving 3m³ to the left gives −2m³ + 3m² − 2 = 0; the highest power of m is 3, so it is not a quadratic equation.
Answer: (1) quadratic; (2) quadratic; (3) not quadratic; (4) quadratic; (5) quadratic; (6) not quadratic
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Practice Set 2.1, Q3
Write each in the form ax² + bx + c = 0 and give a, b, c: (1) 2y = 10 − y² (2) (x − 1)² = 2x + 3 (3) x² + 5x = −(3 − x) (4) 3m² = 2m² − 9 (5) p(3 + 6p) = −5 (6) x² − 9 = 13
1. (1) 2y = 10 − y²
2. y² + 2y − 10 = 0
3. a = 1, b = 2, c = -10
4. (2) (x − 1)² = 2x + 3
5. x² − 2x + 1 = 2x + 3
6. x² − 4x − 2 = 0
7. a = 1, b = -4, c = -2
8. (3) x² + 5x = −(3 − x)
9. x² + 5x = −3 + x
10. x² + 4x + 3 = 0
11. a = 1, b = 4, c = 3
12. (4) 3m² = 2m² − 9
13. m² + 9 = 0
14. a = 1, b = 0, c = 9
15. (5) p(3 + 6p) = −5
16. 3p + 6p² = −5
17. 6p² + 3p + 5 = 0
18. a = 6, b = 3, c = 5
19. (6) x² − 9 = 13
20. x² − 22 = 0
21. a = 1, b = 0, c = -22
Answer: (1) a = 1, b = 2, c = -10; (2) a = 1, b = -4, c = -2; (3) a = 1, b = 4, c = 3; (4) a = 1, b = 0, c = 9; (5) a = 6, b = 3, c = 5; (6) a = 1, b = 0, c = -22
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Practice Set 2.1, Q4
Check whether the given values are roots: (1) x² + 4x − 5 = 0; x = 1, −1 (2) 2m² − 5m = 0; m = 2, 5/2
1. (1) Substitute x = 1 into x² + 4x − 5 = 0
2. 1² + 4(1) − 5 = 1 + 4 − 5 = 0
3. Since LHS = RHS, x = 1 is a root
4. Substitute x = −1 into x² + 4x − 5 = 0
5. (−1)² + 4(−1) − 5 = 1 − 4 − 5 = −8
6. Since LHS ≠ RHS, x = −1 is not a root
7. (2) Substitute m = 2 into 2m² − 5m = 0
8. 2(2)² − 5(2) = 2(4) − 10 = 8 − 10 = −2
9. Since LHS ≠ RHS, m = 2 is not a root
10. Substitute m = 5/2 into 2m² − 5m = 0
11. 2(5/2)² − 5(5/2) = 2(25/4) − 25/2 = 25/2 − 25/2 = 0
12. Since LHS = RHS, m = 5/2 is a root
Answer: (1) 1 is a root; -1 is not a root; (2) 5/2 is a root; 2 is not a root
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Practice Set 2.1, Q5
Find k if x = 3 is a root of kx² − 10x + 3 = 0.
1. Since x = 3 is a root, substitute x = 3 into the equation kx² − 10x + 3 = 0.
2. k(3)² − 10(3) + 3 = 0
3. 9k − 30 + 3 = 0
4. 9k − 27 = 0
5. 9k = 27
6. k = 27 ÷ 9
7. k = 3
Answer: k = 3
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Practice Set 2.1, Q6
One root of 5m² + 2m + k = 0 is −7/5. Find k.
1. Given equation: 5m² + 2m + k = 0
2. One root is m = −7/5
3. Substitute m = −7/5 into the equation
4. 5(−7/5)² + 2(−7/5) + k = 0
5. 5(49/25) − 14/5 + k = 0
6. 49/5 − 14/5 + k = 0
7. 35/5 + k = 0
8. 7 + k = 0
9. k = −7
Answer: k = -7
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Practice Set 2.2

Not solved here yet: Q1(7) (see the textbook).

Practice Set 2.2, Q1(1)
Solve by factorisation: x² − 15x + 54 = 0.
1. Given equation is x² − 15x + 54 = 0.
2. Find two numbers whose product is 54 and sum is −15.
3. The numbers are −9 and −6 because (−9) × (−6) = 54 and (−9) + (−6) = −15.
4. Split the middle term: x² − 9x − 6x + 54 = 0.
5. Group terms: x(x − 9) − 6(x − 9) = 0.
6. Factor out the common binomial: (x − 9)(x − 6) = 0.
7. Set each factor to zero: x − 9 = 0 or x − 6 = 0.
8. Solve for x: x = 9 or x = 6.
Answer: x = 9 or x = 6
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Practice Set 2.2, Q1(2)
Solve by factorisation: x² + x − 20 = 0.
1. x² + x − 20 = 0
2. x² + 5x − 4x − 20 = 0
3. x(x + 5) − 4(x + 5) = 0
4. (x + 5)(x − 4) = 0
5. x + 5 = 0 or x − 4 = 0
6. x = −5 or x = 4
Answer: x = −5 or x = 4
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Practice Set 2.2, Q1(3)
Solve by factorisation: 2y² + 27y + 13 = 0.
1. 2y² + 27y + 13 = 0
2. Split the middle term 27y into 26y + y
3. 2y² + 26y + y + 13 = 0
4. Group terms: 2y(y + 13) + 1(y + 13) = 0
5. Factor out the common binomial: (2y + 1)(y + 13) = 0
6. Set each factor to zero: 2y + 1 = 0 or y + 13 = 0
7. From 2y + 1 = 0, we get 2y = -1
8. y = -1/2
9. From y + 13 = 0, we get y = -13
Answer: y = -13; y = -1/2
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Practice Set 2.2, Q1(4)
Solve by factorisation: 5m² = 22m + 15.
1. 5m² − 22m − 15 = 0
2. Product of 5 and −15 is −75; sum is −22
3. Split middle term: 5m² − 25m + 3m − 15 = 0
4. Group terms: 5m(m − 5) + 3(m − 5) = 0
5. Factor out common binomial: (5m + 3)(m − 5) = 0
6. Set each factor to zero: 5m + 3 = 0 or m − 5 = 0
7. Solve: m = −3/5 or m = 5
Answer: m = 5 or m = −3/5
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Practice Set 2.2, Q1(5)
Solve by factorisation: 2x² − 2x + 1/2 = 0.
1. 2x² − 2x + 1/2 = 0
2. Multiply both sides by 2 to clear the fraction: 4x² − 4x + 1 = 0
3. Recognise the left side as a perfect square: (2x − 1)² = 0
4. Take the square root of both sides: 2x − 1 = 0
5. Add 1 to both sides: 2x = 1
6. Divide by 2: x = 1/2
Answer: x = 1/2
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Practice Set 2.2, Q1(6)
Solve by factorisation: 6x − 2/x = 1.
1. 6x − 2/x = 1
2. Multiply both sides by x to clear the denominator: 6x² − 2 = x
3. Rearrange into standard form: 6x² − x − 2 = 0
4. Split the middle term using 2x and −3x: 6x² + 2x − 3x − 2 = 0
5. Group and factor: 2x(3x + 1) − 1(3x + 2) = 0
6. Factor by grouping: (3x + 1)(2x − 1) = 0
7. Set each factor to zero: 3x + 1 = 0 or 2x − 1 = 0
8. Solve 3x + 1 = 0: 3x = −1, so x = −1/3
9. Solve 2x − 1 = 0: 2x = 1, so x = 1/2
10. Check x = −1/3: 6(−1/3) − 2/(−1/3) = −2 + 6 = 4 ≠ 1, so x = −1/3 is extraneous
11. Check x = 1/2: 6(1/2) − 2/(1/2) = 3 − 4 = −1 ≠ 1, so x = 1/2 is extraneous
12. Re-evaluate factorisation: 6x² − x − 2 = 0, discriminant = (−1)² − 4(6)(−2) = 1 + 48 = 49
13. Roots: x = (1 ± 7)/12, so x = 8/12 = 2/3 or x = −6/12 = −1/2
14. Check x = 2/3: 6(2/3) − 2/(2/3) = 4 − 3 = 1, correct
15. Check x = −1/2: 6(−1/2) − 2/(−1/2) = −3 + 4 = 1, correct
Answer: x = 2/3 or x = −1/2
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Practice Set 2.2, Q1(8)
Solve by factorisation: 3x² − 2√6x + 2 = 0.
1. 3x² − 2√6x + 2 = 0
2. Split the middle term: 3x² − √6x − √6x + 2 = 0
3. Group terms: x(3x − √6) − √6(x − √6/3) = 0
4. Factor out: (3x − √6)(x − √6/3) = 0
5. Set each factor to zero: 3x − √6 = 0 or x − √6/3 = 0
6. From 3x − √6 = 0, 3x = √6, so x = √6/3
7. From x − √6/3 = 0, x = √6/3
8. Simplify √6/3 = √(2×3)/√(3×3) = √2/√3
9. Thus x = √2/√3
Answer: x = √2/√3
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Practice Set 2.2, Q1(9)
Solve by factorisation: 2m(m − 24) = 50.
1. 2m(m − 24) = 50
2. 2m² − 48m = 50
3. 2m² − 48m − 50 = 0
4. Divide by 2: m² − 24m − 25 = 0
5. Split middle term: m² − 25m + m − 25 = 0
6. m(m − 25) + 1(m − 25) = 0
7. (m + 1)(m − 25) = 0
8. m + 1 = 0 or m − 25 = 0
9. m = −1 or m = 25
Answer: m = 25; m = −1
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Practice Set 2.2, Q1(10)
Solve by factorisation: 25m² = 9.
1. 25m² = 9
2. 25m² − 9 = 0
3. (5m)² − (3)² = 0
4. (5m − 3)(5m + 3) = 0
5. 5m − 3 = 0 or 5m + 3 = 0
6. 5m = 3 or 5m = −3
7. m = 3/5 or m = −3/5
Answer: m = 3/5 or m = −3/5
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Practice Set 2.2, Q1(11)
Solve by factorisation: 7m² = 21m.
1. Given equation: 7m² = 21m
2. Move all terms to one side: 7m² − 21m = 0
3. Factor out the common term 7m: 7m(m − 3) = 0
4. Set each factor equal to zero: 7m = 0 or m − 3 = 0
5. Solve for m: m = 0 or m = 3
Answer: m = 0 or m = 3
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Practice Set 2.2, Q1(12)
Solve by factorisation: m² − 11 = 0.
1. Given equation is m² − 11 = 0.
2. Move the constant to the right side: m² = 11.
3. Take the square root of both sides: m = ±√11.
4. The two roots are m = √11 and m = −√11.
Answer: m = √11, m = −√11
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Practice Set 2.3

Practice Set 2.3, Q1(1)
Solve by completing the square: x² + x − 20 = 0.
1. Given equation: x² + x − 20 = 0
2. Move the constant to the right side: x² + x = 20
3. Add (1/2)² = 1/4 to both sides: x² + x + 1/4 = 20 + 1/4
4. Simplify the right side: 20 + 1/4 = 81/4
5. Write the left side as a perfect square: (x + 1/2)² = 81/4
6. Take the square root of both sides: x + 1/2 = ±√(81/4)
7. Simplify the square root: √(81/4) = 9/2
8. So, x + 1/2 = 9/2 or x + 1/2 = −9/2
9. Solve for x in the first case: x = 9/2 − 1/2 = 8/2 = 4
10. Solve for x in the second case: x = −9/2 − 1/2 = −10/2 = −5
Answer: x = 4 or x = −5
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Practice Set 2.3, Q1(2)
Solve by completing the square: x² + 2x − 5 = 0.
1. x² + 2x − 5 = 0
2. x² + 2x = 5
3. Add (2/2)² = 1 to both sides
4. x² + 2x + 1 = 5 + 1
5. (x + 1)² = 6
6. x + 1 = ±√6
7. x = −1 + √6 or x = −1 − √6
Answer: x = -1 + √6 or x = -1 - √6
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Practice Set 2.3, Q1(3)
Solve by completing the square: m² − 5m = −3.
1. m² − 5m = −3
2. m² − 5m + (5/2)² = −3 + (5/2)²
3. m² − 5m + 25/4 = −3 + 25/4
4. (m − 5/2)² = 13/4
5. m − 5/2 = ±√(13/4)
6. m − 5/2 = ±√13/2
7. m = 5/2 ± √13/2
8. m = (5 + √13)/2 or m = (5 − √13)/2
Answer: m = (5 + √13)/2 or m = (5 - √13)/2
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Practice Set 2.3, Q1(4)
Solve by completing the square: 9y² − 12y + 2 = 0.
1. 9y² − 12y + 2 = 0
2. Divide the entire equation by 9: y² − (4/3)y + 2/9 = 0
3. Move the constant term to the right side: y² − (4/3)y = −2/9
4. Take half of the coefficient of y, which is −4/3, giving −2/3
5. Square this value: (−2/3)² = 4/9
6. Add 4/9 to both sides: y² − (4/3)y + 4/9 = −2/9 + 4/9
7. Simplify the right side: y² − (4/3)y + 4/9 = 2/9
8. Write the left side as a perfect square: (y − 2/3)² = 2/9
9. Take the square root of both sides: y − 2/3 = ±√(2/9)
10. Simplify the square root: y − 2/3 = ±√2/3
11. Add 2/3 to both sides: y = 2/3 ± √2/3
12. Combine the fractions: y = (2 ± √2)/3
Answer: (2 + √2)/3, (2 - √2)/3
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Practice Set 2.3, Q1(5)
Solve by completing the square: 2y² + 9y + 10 = 0.
1. 2y² + 9y + 10 = 0
2. Divide the equation by 2 to make the coefficient of y² equal to 1: y² + (9/2)y + 5 = 0
3. Move the constant term to the right side: y² + (9/2)y = -5
4. Take half of the coefficient of y, which is (9/2) ÷ 2 = 9/4
5. Square this value: (9/4)² = 81/16
6. Add 81/16 to both sides of the equation: y² + (9/2)y + 81/16 = -5 + 81/16
7. Simplify the right side: -5 + 81/16 = -80/16 + 81/16 = 1/16
8. Write the left side as a perfect square: (y + 9/4)² = 1/16
9. Take the square root of both sides: y + 9/4 = ±√(1/16)
10. Simplify the square root: y + 9/4 = ±1/4
11. Case 1: y + 9/4 = 1/4, so y = 1/4 - 9/4 = -8/4 = -2
12. Case 2: y + 9/4 = -1/4, so y = -1/4 - 9/4 = -10/4 = -5/2
Answer: y = -2 or y = -5/2
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Practice Set 2.3, Q1(6)
Solve by completing the square: 5x² = 4x + 7.
1. 5x² − 4x − 7 = 0
2. x² − (4/5)x − (7/5) = 0
3. x² − (4/5)x = 7/5
4. x² − (4/5)x + (2/5)² = 7/5 + (2/5)²
5. (x − 2/5)² = 7/5 + 4/25
6. (x − 2/5)² = 35/25 + 4/25
7. (x − 2/5)² = 39/25
8. x − 2/5 = ±√(39/25)
9. x − 2/5 = ±√39/5
10. x = 2/5 ± √39/5
11. x = (2 + √39)/5 or x = (2 − √39)/5
Answer: x = (2 + √39)/5 or x = (2 − √39)/5
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Practice Set 2.4

Practice Set 2.4, Q1
Compare with ax² + bx + c = 0 and write a, b, c: (1) x² − 7x + 5 = 0 (2) 2m² = 5m − 5 (3) y² = 7y
1. (1) The given equation is x² − 7x + 5 = 0.
2. (1) Comparing it with ax² + bx + c = 0, we see a = 1, b = −7, c = 5.
3. (2) The given equation is 2m² = 5m − 5.
4. (2) Rearranging to standard form gives 2m² − 5m + 5 = 0.
5. (2) Comparing with ax² + bx + c = 0, we see a = 2, b = −5, c = 5.
6. (3) The given equation is y² = 7y.
7. (3) Rearranging to standard form gives y² − 7y = 0.
8. (3) Comparing with ax² + bx + c = 0, we see a = 1, b = −7, c = 0.
Answer: (1) a = 1, b = -7, c = 5; (2) a = 2, b = -5, c = 5; (3) a = 1, b = -7, c = 0
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Practice Set 2.4, Q2(1)
Solve using the formula: x² + 6x + 5 = 0.
1. Compare the equation x² + 6x + 5 = 0 with ax² + bx + c = 0 to get a = 1, b = 6, c = 5.
2. Write the quadratic formula: x = (−b ± √(b² − 4ac)) / (2a).
3. Calculate the discriminant: b² − 4ac = 6² − 4 × 1 × 5 = 36 − 20 = 16.
4. Substitute the values into the formula: x = (−6 ± √16) / (2 × 1).
5. Simplify the square root and denominator: x = (−6 ± 4) / 2.
6. Find the first root: x = (−6 + 4) / 2 = −2 / 2 = −1.
7. Find the second root: x = (−6 − 4) / 2 = −10 / 2 = −5.
Answer: x = −1 or x = −5
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Practice Set 2.4, Q2(2)
Solve using the formula: x² − 3x − 2 = 0.
1. Given equation: x² − 3x − 2 = 0
2. Comparing with ax² + bx + c = 0, we get a = 1, b = −3, c = −2
3. Discriminant D = b² − 4ac = (−3)² − 4(1)(−2)
4. D = 9 + 8 = 17
5. Since D > 0, the roots are real and distinct
6. Using the formula x = (−b ± √D) / 2a
7. x = (3 ± √17) / 2
8. Therefore, x = (3 + √17)/2 or x = (3 − √17)/2
Answer: x = (3 + √17)/2, (3 - √17)/2
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Practice Set 2.4, Q2(3)
Solve using the formula: 3m² + 2m − 7 = 0.
1. Given equation is 3m² + 2m − 7 = 0
2. Compare with am² + bm + c = 0 to get a = 3, b = 2, c = −7
3. Formula for m is m = (−b ± √(b² − 4ac)) / 2a
4. Substitute values to get m = (−2 ± √(2² − 4 × 3 × (−7))) / (2 × 3)
5. Calculate inside the square root: 2² − 4 × 3 × (−7) = 4 + 84 = 88
6. Simplify the expression to m = (−2 ± √88) / 6
7. Simplify √88 as √(4 × 22) = 2√22
8. Substitute back to get m = (−2 ± 2√22) / 6
9. Factor out 2 from the numerator to get m = 2(−1 ± √22) / 6
10. Divide numerator and denominator by 2 to get m = (−1 ± √22) / 3
11. Split into two cases: m = (−1 + √22) / 3 or m = (−1 − √22) / 3
Answer: m = (-1 + √22)/3 or m = (-1 - √22)/3
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Practice Set 2.4, Q2(4)
Solve using the formula: 5m² − 4m − 2 = 0.
1. Compare 5m² − 4m − 2 = 0 with am² + bm + c = 0 to get a = 5, b = −4, c = −2.
2. Calculate the discriminant D = b² − 4ac = (−4)² − 4(5)(−2).
3. Simplify D = 16 + 40 = 56.
4. Apply the quadratic formula m = (−b ± √D) / 2a.
5. Substitute values to get m = (4 ± √56) / 10.
6. Simplify √56 as √(4 × 14) = 2√14.
7. Substitute back to get m = (4 ± 2√14) / 10.
8. Divide numerator and denominator by 2 to get m = (2 ± √14) / 5.
9. Write the two roots as m = (2 + √14)/5 and m = (2 − √14)/5.
Answer: m = (2 + √14)/5, m = (2 - √14)/5
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Practice Set 2.4, Q2(5)
Solve using the formula: y² + (1/3)y = 2.
1. Multiply the equation by 3 to remove the fraction: 3y² + y = 6
2. Write the equation in standard form: 3y² + y − 6 = 0
3. Identify the coefficients: a = 3, b = 1, c = −6
4. Calculate the discriminant: b² − 4ac = 1² − 4(3)(−6)
5. Simplify the discriminant: 1 + 72 = 73
6. Apply the quadratic formula: y = (−b ± √(b² − 4ac)) / 2a
7. Substitute the values: y = (−1 ± √73) / 6
8. The two solutions are y = (−1 + √73)/6 and y = (−1 − √73)/6
Answer: y = (-1 + √73)/6, y = (-1 - √73)/6
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Practice Set 2.4, Q2(6)
Solve using the formula: 5x² + 13x + 8 = 0.
1. Given equation is 5x² + 13x + 8 = 0
2. Compare with ax² + bx + c = 0 to get a = 5, b = 13, c = 8
3. Calculate discriminant D = b² − 4ac
4. D = 13² − 4 × 5 × 8
5. D = 169 − 160
6. D = 9
7. Use formula x = (−b ± √D) / 2a
8. x = (−13 ± √9) / (2 × 5)
9. x = (−13 ± 3) / 10
10. First root x₁ = (−13 + 3) / 10
11. x₁ = −10 / 10
12. x₁ = −1
13. Second root x₂ = (−13 − 3) / 10
14. x₂ = −16 / 10
15. x₂ = −8/5
Answer: x = −1, −8/5
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Practice Set 2.4, Q3
Solve x² + 2√3x + 3 = 0 using the formula (find a, b, c, then b² − 4ac, then the roots).
1. Identify the coefficients from the equation x² + 2√3x + 3 = 0
2. a = 1, b = 2√3, c = 3
3. Calculate the discriminant D = b² − 4ac
4. D = (2√3)² − 4(1)(3)
5. D = 12 − 12
6. D = 0
7. Since D = 0, the roots are real and equal
8. Use the formula x = −b / 2a
9. x = −(2√3) / (2 × 1)
10. x = −2√3 / 2
11. x = −√3
Answer: x = −√3 (the two roots are equal)
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Practice Set 2.5

Practice Set 2.5, Q1
(1) What is the nature of the roots when b² − 4ac = 5, and when b² − 4ac = −5? (2) Write the quadratic equation whose roots have sum −7 and product 5. (3) For 2x² − 4x − 3 = 0, find α + β and α × β.
1. (1) For b² − 4ac = 5, the discriminant is positive.
2. When the discriminant is positive, the roots are real and unequal.
3. (1) For b² − 4ac = −5, the discriminant is negative.
4. When the discriminant is negative, the roots are not real.
5. (2) The standard form of a quadratic equation with roots α and β is x² − (α + β)x + αβ = 0.
6. (2) Given sum of roots α + β = −7 and product αβ = 5.
7. (2) Substituting these values gives x² − (−7)x + 5 = 0.
8. (2) Simplifying the equation results in x² + 7x + 5 = 0.
9. (3) The given equation is 2x² − 4x − 3 = 0.
10. (3) Comparing with ax² + bx + c = 0, we have a = 2, b = −4, and c = −3.
11. (3) The sum of roots α + β is given by −b/a.
12. (3) Calculating the sum: α + β = −(−4)/2 = 4/2 = 2.
13. (3) The product of roots α × β is given by c/a.
14. (3) Calculating the product: α × β = −3/2.
Answer: (1) real and unequal; not real; (2) x² + 7x + 5 = 0; (3) α + β = 2; α × β = -3/2
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Practice Set 2.5, Q2
Find the discriminant: (1) x² + 7x − 1 = 0 (2) 2y² − 5y + 10 = 0 (3) √2x² + 4x + 2√2 = 0
1. (1) The equation is x² + 7x − 1 = 0, so a = 1, b = 7, c = −1.
2. (1) Discriminant D = b² − 4ac.
3. (1) D = 7² − 4(1)(−1).
4. (1) D = 49 + 4.
5. (1) D = 53.
6. (2) The equation is 2y² − 5y + 10 = 0, so a = 2, b = −5, c = 10.
7. (2) Discriminant D = b² − 4ac.
8. (2) D = (−5)² − 4(2)(10).
9. (2) D = 25 − 80.
10. (2) D = −55.
11. (3) The equation is √2x² + 4x + 2√2 = 0, so a = √2, b = 4, c = 2√2.
12. (3) Discriminant D = b² − 4ac.
13. (3) D = 4² − 4(√2)(2√2).
14. (3) D = 16 − 4(2 × 2).
15. (3) D = 16 − 16.
16. (3) D = 0.
Answer: (1) 53; (2) -55; (3) 0
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Practice Set 2.5, Q3
Find the nature of the roots: (1) x² − 4x + 4 = 0 (2) 2y² − 7y + 2 = 0 (3) m² + 2m + 9 = 0
1. (1) a = 1, b = −4, c = 4
2. (1) D = b² − 4ac = (−4)² − 4(1)(4)
3. (1) D = 16 − 16 = 0
4. (1) Since D = 0, roots are real and equal
5. (2) a = 2, b = −7, c = 2
6. (2) D = b² − 4ac = (−7)² − 4(2)(2)
7. (2) D = 49 − 16 = 33
8. (2) Since D > 0, roots are real and unequal
9. (3) a = 1, b = 2, c = 9
10. (3) D = b² − 4ac = (2)² − 4(1)(9)
11. (3) D = 4 − 36 = −32
12. (3) Since D < 0, roots are not real
Answer: (1) real and equal; (2) real and unequal; (3) not real
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Practice Set 2.5, Q4
Form the quadratic equation whose roots are: (1) 0 and 4 (2) 3 and −10 (3) 1/2 and −1/2 (4) 2 − √5 and 2 + √5
1. (1) Sum of roots = 0 + 4 = 4
2. (1) Product of roots = 0 × 4 = 0
3. (1) Equation: x² − (sum)x + (product) = 0
4. (1) x² − 4x + 0 = 0
5. (1) x² - 4x = 0
6. (2) Sum of roots = 3 + (−10) = −7
7. (2) Product of roots = 3 × (−10) = −30
8. (2) Equation: x² − (−7)x + (−30) = 0
9. (2) x² + 7x − 30 = 0
10. (3) Sum of roots = 1/2 + (−1/2) = 0
11. (3) Product of roots = (1/2) × (−1/2) = −1/4
12. (3) Equation: x² − 0x + (−1/4) = 0
13. (3) x² − 1/4 = 0
14. (4) Sum of roots = (2 − √5) + (2 + √5) = 4
15. (4) Product of roots = (2 − √5)(2 + √5) = 4 − 5 = −1
16. (4) Equation: x² − 4x + (−1) = 0
17. (4) x² − 4x − 1 = 0
Answer: (1) x² - 4x = 0; (2) x² + 7x - 30 = 0; (3) x² - 1/4 = 0; (4) x² - 4x - 1 = 0
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Practice Set 2.5, Q5
For x² − 4kx + k + 3 = 0 the sum of the roots is twice their product. Find k.
1. Given equation: x² − 4kx + (k + 3) = 0
2. Compare with ax² + bx + c = 0 to get a = 1, b = −4k, c = k + 3
3. Sum of roots = −b/a = −(−4k)/1 = 4k
4. Product of roots = c/a = (k + 3)/1 = k + 3
5. Condition: Sum of roots = 2 × Product of roots
6. 4k = 2(k + 3)
7. 4k = 2k + 6
8. 4k − 2k = 6
9. 2k = 6
10. k = 6/2
11. k = 3
Answer: k = 3
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Practice Set 2.5, Q6
α and β are the roots of y² − 2y − 7 = 0. Find (1) α² + β² (2) α³ + β³.
1. Given equation is y² − 2y − 7 = 0.
2. Comparing with ay² + by + c = 0, we get a = 1, b = −2, c = −7.
3. Sum of roots α + β = −b/a = −(−2)/1 = 2.
4. Product of roots αβ = c/a = −7/1 = −7.
5. (1) α² + β² = (α + β)² − 2αβ.
6. α² + β² = (2)² − 2(−7).
7. α² + β² = 4 + 14.
8. α² + β² = 18.
9. (2) α³ + β³ = (α + β)³ − 3αβ(α + β).
10. α³ + β³ = (2)³ − 3(−7)(2).
11. α³ + β³ = 8 − (−42).
12. α³ + β³ = 8 + 42.
13. α³ + β³ = 50.
Answer: (1) 18; (2) 50
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Practice Set 2.5, Q7
Find k so that the roots are real and equal: (1) 3y² + ky + 12 = 0 (2) kx(x − 2) + 6 = 0
1. (1) The quadratic equation is 3y² + ky + 12 = 0.
2. Here, a = 3, b = k, and c = 12.
3. For real and equal roots, the discriminant D must be zero.
4. D = b² − 4ac = k² − 4(3)(12).
5. k² − 144 = 0.
6. k² = 144.
7. k = 12 or k = -12.
8. (2) The equation is kx(x − 2) + 6 = 0.
9. Expanding it gives kx² − 2kx + 6 = 0.
10. Here, a = k, b = -2k, and c = 6.
11. For real and equal roots, D = 0.
12. D = (-2k)² − 4(k)(6).
13. 4k² − 24k = 0.
14. 4k(k − 6) = 0.
15. Since a ≠ 0 for a quadratic equation, k ≠ 0.
16. Therefore, k − 6 = 0.
17. k = 6.
Answer: (1) k = 12, k = -12; (2) k = 6
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Practice Set 2.6

Practice Set 2.6, Q1
The product of Pragati's age 2 years ago and her age 3 years from now is 84. Find her present age.
1. Let Pragati's present age be x years.
2. Her age 2 years ago is (x − 2) years.
3. Her age 3 years from now is (x + 3) years.
4. According to the problem, the product of these ages is 84.
5. (x − 2)(x + 3) = 84
6. x² + 3x − 2x − 6 = 84
7. x² + x − 6 = 84
8. x² + x − 90 = 0
9. Split the middle term: x² + 10x − 9x − 90 = 0
10. x(x + 10) − 9(x + 10) = 0
11. (x + 10)(x − 9) = 0
12. So, x + 10 = 0 or x − 9 = 0
13. x = −10 or x = 9
14. Age cannot be negative, so x = 9.
Answer: Pragati's present age is 9 years.
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Practice Set 2.6, Q2
The squares of two consecutive even natural numbers add up to 244. Find the numbers.
1. Let the two consecutive even natural numbers be x and x + 2.
2. According to the problem, the sum of their squares is 244.
3. x² + (x + 2)² = 244
4. x² + x² + 4x + 4 = 244
5. 2x² + 4x + 4 - 244 = 0
6. 2x² + 4x - 240 = 0
7. Divide the entire equation by 2 to simplify.
8. x² + 2x - 120 = 0
9. Split the middle term 2x into 12x and -10x.
10. x² + 12x - 10x - 120 = 0
11. x(x + 12) - 10(x + 12) = 0
12. (x + 12)(x - 10) = 0
13. x + 12 = 0 or x - 10 = 0
14. x = -12 or x = 10
15. Since the numbers are natural, x cannot be negative.
16. Therefore, x = 10.
17. The first number is 10.
18. The second number is x + 2 = 10 + 2 = 12.
Answer: 10 and 12
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Practice Set 2.6, Q3
A garden has 150 orange trees; each row has 5 more trees than each column. Find the number of trees in each column and in each row.
1. Let the number of trees in each column be x.
2. Then the number of trees in each row is x + 5.
3. The total number of trees is the product of trees in a row and trees in a column.
4. x(x + 5) = 150
5. x² + 5x = 150
6. x² + 5x − 150 = 0
7. Split the middle term: 15x − 10x
8. x² + 15x − 10x − 150 = 0
9. x(x + 15) − 10(x + 15) = 0
10. (x + 15)(x − 10) = 0
11. x + 15 = 0 or x − 10 = 0
12. x = −15 or x = 10
13. Number of trees cannot be negative, so x = 10.
14. Number of trees in each column is 10.
15. Number of trees in each row is 10 + 5 = 15.
Answer: Number of trees in each column = 10; Number of trees in each row = 15
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Practice Set 2.6, Q4
Vivek is 5 years older than Kishor, and the reciprocals of their ages add up to 1/6. Find their present ages.
1. Let the present age of Kishor be x years.
2. Then the present age of Vivek is (x + 5) years.
3. According to the problem, the sum of the reciprocals of their ages is 1/6.
4. 1/x + 1/(x + 5) = 1/6
5. (x + 5 + x) / [x(x + 5)] = 1/6
6. (2x + 5) / (x² + 5x) = 1/6
7. 6(2x + 5) = x² + 5x
8. 12x + 30 = x² + 5x
9. x² + 5x - 12x - 30 = 0
10. x² - 7x - 30 = 0
11. x² - 10x + 3x - 30 = 0
12. x(x - 10) + 3(x - 10) = 0
13. (x - 10)(x + 3) = 0
14. x = 10 or x = -3
15. Since age cannot be negative, x = 10.
16. Kishor's age is 10 years.
17. Vivek's age is 10 + 5 = 15 years.
Answer: Kishor 10; Vivek 15
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Practice Set 2.6, Q5
Suyash scored 10 marks more in the second test than in the first, and 5 times the second score equals the square of the first score. Find his score in the first test.
1. Let Suyash's score in the first test be x.
2. Then his score in the second test is x + 10.
3. According to the condition, 5 times the second score equals the square of the first score.
4. 5(x + 10) = x²
5. 5x + 50 = x²
6. x² − 5x − 50 = 0
7. x² − 10x + 5x − 50 = 0
8. x(x − 10) + 5(x − 10) = 0
9. (x − 10)(x + 5) = 0
10. x = 10 or x = −5
11. Since a test score cannot be negative, x = 10.
12. Therefore, Suyash's score in the first test is 10.
Answer: 10
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Practice Set 2.6, Q6
A potter makes some pots a day. The cost of making one pot is ₹40 more than 10 times the number of pots made that day, and the day's total cost is ₹600. Find the number of pots and the cost of one pot.
1. Let the number of pots made in a day be x.
2. The cost of making one pot is 10x + 40.
3. The total cost is the product of the number of pots and the cost per pot, so x(10x + 40) = 600.
4. Expanding the equation gives 10x² + 40x = 600.
5. Dividing the entire equation by 10 simplifies it to x² + 4x = 60.
6. Rearranging the terms gives x² + 4x − 60 = 0.
7. Factorizing the quadratic expression gives (x + 10)(x − 6) = 0.
8. Setting each factor to zero gives x + 10 = 0 or x − 6 = 0.
9. Solving for x gives x = −10 or x = 6.
10. Since the number of pots cannot be negative, we take x = 6.
11. The cost of one pot is 10(6) + 40.
12. This simplifies to 60 + 40 = 100.
Answer: 6 pots; ₹100
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Practice Set 2.6, Q7
A boat goes 36 km downstream and back in 8 hours in all. Its speed in still water is 12 km/h. Find the speed of the current.
1. Let the speed of the current be x km/h.
2. Speed downstream = (12 + x) km/h and speed upstream = (12 − x) km/h.
3. Time taken downstream = 36 ÷ (12 + x) hours.
4. Time taken upstream = 36 ÷ (12 − x) hours.
5. Total time is given as 8 hours, so 36 ÷ (12 + x) + 36 ÷ (12 − x) = 8.
6. Multiply both sides by (12 + x)(12 − x) to clear denominators: 36(12 − x) + 36(12 + x) = 8(12 + x)(12 − x).
7. Simplify left side: 432 − 36x + 432 + 36x = 864.
8. Simplify right side: 8(144 − x²) = 1152 − 8x².
9. Equation becomes 864 = 1152 − 8x².
10. Rearrange: 8x² = 1152 − 864.
11. 8x² = 288.
12. x² = 288 ÷ 8.
13. x² = 36.
14. x = 6 or x = −6
15. The speed of the current cannot be negative, so x = 6 (and 6 < 12, so the boat can still go upstream).
Answer: Speed of the current = 6 km/h
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Practice Set 2.6, Q8
Pintu takes 6 days more than Nishu to finish a work alone; together they finish it in 4 days. How many days does each take alone?
1. Let Nishu take x days to finish the work alone.
2. Then Pintu takes (x + 6) days to finish the work alone.
3. Nishu's one day work is 1/x and Pintu's one day work is 1/(x + 6).
4. Together, their one day work is 1/4.
5. So, 1/x + 1/(x + 6) = 1/4.
6. Taking LCM of denominators, (x + 6 + x) / [x(x + 6)] = 1/4.
7. Simplifying the numerator, (2x + 6) / [x(x + 6)] = 1/4.
8. Cross multiplying, 4(2x + 6) = x(x + 6).
9. Expanding both sides, 8x + 24 = x² + 6x.
10. Rearranging terms to form a quadratic equation, x² + 6x - 8x - 24 = 0.
11. Simplifying, x² - 2x - 24 = 0.
12. Factoring the quadratic equation, x² - 6x + 4x - 24 = 0.
13. Grouping terms, x(x - 6) + 4(x - 6) = 0.
14. Taking common factor, (x - 6)(x + 4) = 0.
15. So, x - 6 = 0 or x + 4 = 0.
16. Thus, x = 6 or x = -4.
17. Since time cannot be negative, x = 6.
18. Therefore, Nishu takes 6 days to finish the work alone.
19. Pintu takes x + 6 = 6 + 6 = 12 days to finish the work alone.
Answer: Nishu 6 days; Pintu 12 days
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Practice Set 2.6, Q9
When 460 is divided by a natural number, the quotient is 6 more than five times the divisor and the remainder is 1. Find the quotient and the divisor.
1. Let the divisor be x.
2. The quotient is 5x + 6.
3. Using the division algorithm, 460 = x(5x + 6) + 1.
4. 460 = 5x² + 6x + 1.
5. 5x² + 6x + 1 - 460 = 0.
6. 5x² + 6x - 459 = 0.
7. Using the formula, x = (−6 ± √(36 + 9180)) / 10, and the divisor is a natural number, so take the + sign: x = (−6 + √(36 + 9180)) / 10.
8. x = (-6 + √9216) / 10.
9. x = (-6 + 96) / 10.
10. x = 90 / 10.
11. x = 9.
12. The divisor is 9.
13. The quotient is 5(9) + 6.
14. Quotient = 45 + 6.
15. Quotient = 51.
Answer: divisor 9; quotient 51
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Practice Set 2.6, Q10
Trapezium ABCD with AB ∥ CD has area 33 cm². From the figure: AB = x, CD = 2x + 1, the height AM = x − 4 and AD = BC = x − 2. Find the lengths of all the sides.
Uses a figure printed in the textbook — open the PDF above to see it.
1. Area of trapezium ABCD = 33 cm²
2. Formula for area = 1/2 × (sum of parallel sides) × height
3. Substitute given values: 33 = 1/2 × (AB + CD) × AM
4. 33 = 1/2 × (x + 2x + 1) × (x − 4)
5. 33 = 1/2 × (3x + 1) × (x − 4)
6. Multiply both sides by 2: 66 = (3x + 1)(x − 4)
7. Expand the right side: 66 = 3x² − 12x + x − 4
8. Simplify: 66 = 3x² − 11x − 4
9. Rearrange to form quadratic equation: 3x² − 11x − 70 = 0
10. Factorize: 3x² − 21x + 10x − 70 = 0
11. Group terms: 3x(x − 7) + 10(x − 7) = 0
12. Factor out common term: (3x + 10)(x − 7) = 0
13. Solutions are x = −10/3 or x = 7
14. Since length cannot be negative, x = 7
15. Calculate AB = x = 7 cm
16. Calculate CD = 2x + 1 = 2(7) + 1 = 15 cm
17. Calculate AD = BC = x − 2 = 7 − 2 = 5 cm
Answer: AB = 7 cm; CD = 15 cm; AD = BC = 5 cm
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Problem Set 2

Not solved here yet: Q10 (see the textbook).

Problem Set 2, Q1(1)
Choose the correct option: which one is a quadratic equation? (A) 5/x − 3 = x² (B) x(x + 5) = 2 (C) n − 1 = 2n (D) (1/x²)(x + 2) = x
1. A quadratic equation must be of the form ax² + bx + c = 0 where a ≠ 0 and the variable is in the numerator.
2. Option (A) has x in the denominator, so it is not a polynomial equation.
3. Option (C) simplifies to n − 1 = 2n, which is a linear equation.
4. Option (D) has x² in the denominator, so it is not a polynomial equation.
5. Option (B) expands to x² + 5x = 2, which rearranges to x² + 5x − 2 = 0.
6. This is a quadratic equation in standard form.
7. (B)
Answer: (B)
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Problem Set 2, Q1(2)
Choose the correct option: which one is not a quadratic equation? (A) x² + 4x = 11 + x² (B) x² = 4x (C) 5x² = 90 (D) 2x − x² = x² + 5
1. (A) x² + 4x = 11 + x²
2. Subtract x² from both sides: 4x = 11
3. The highest power of x is 1, so it is a linear equation, not quadratic
4. (B) x² = 4x
5. Rearrange: x² − 4x = 0
6. The highest power of x is 2, so it is a quadratic equation
7. (C) 5x² = 90
8. Rearrange: 5x² − 90 = 0
9. The highest power of x is 2, so it is a quadratic equation
10. (D) 2x − x² = x² + 5
11. Rearrange: −2x² + 2x − 5 = 0
12. The highest power of x is 2, so it is a quadratic equation
13. Option (A) is the only one that is not a quadratic equation
Answer: (A)
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Problem Set 2, Q1(3)
Choose the correct option: the roots of x² + kx + k = 0 are real and equal; k = (A) 0 (B) 4 (C) 0 or 4 (D) 2
1. For roots to be real and equal, the discriminant must be zero.
2. The discriminant D is given by b² − 4ac.
3. Here a = 1, b = k, and c = k.
4. Substitute these values into the formula: D = k² − 4(1)(k).
5. Set the discriminant to zero: k² − 4k = 0.
6. Factor the equation: k(k − 4) = 0.
7. This gives two possible values: k = 0 or k = 4.
8. Therefore, the correct option is (C).
Answer: (C)
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Problem Set 2, Q1(4)
Choose the correct option: the discriminant of √2x² − 5x + √2 = 0 is (A) −5 (B) 17 (C) √2 (D) 2√2 − 5
1. The given quadratic equation is √2x² − 5x + √2 = 0.
2. Comparing with ax² + bx + c = 0, we get a = √2, b = −5, c = √2.
3. The discriminant D is given by the formula b² − 4ac.
4. Substituting the values, D = (−5)² − 4(√2)(√2).
5. Calculating the terms, D = 25 − 4(2).
6. Simplifying, D = 25 − 8.
7. Therefore, D = 17.
8. This matches option (B).
Answer: (B) 17
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Problem Set 2, Q1(5)
Choose the correct option: which equation has roots 3 and 5? (A) x² − 15x + 8 = 0 (B) x² − 8x + 15 = 0 (C) x² + 3x + 5 = 0 (D) x² + 8x − 15 = 0
1. Sum of roots = 3 + 5 = 8
2. Product of roots = 3 × 5 = 15
3. Quadratic equation is x² − (sum)x + (product) = 0
4. Substitute values: x² − 8x + 15 = 0
5. This matches option (B)
Answer: (B)
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Problem Set 2, Q1(6)
Choose the correct option: which equation has sum of roots −5? (A) 3x² − 15x + 3 = 0 (B) x² − 5x + 3 = 0 (C) x² + 3x − 5 = 0 (D) 3x² + 15x + 3 = 0
1. For a quadratic equation ax² + bx + c = 0, the sum of the roots is given by −b/a.
2. Check option (A): a = 3, b = −15, so sum = −(−15)/3 = 15/3 = 5.
3. Check option (B): a = 1, b = −5, so sum = −(−5)/1 = 5/1 = 5.
4. Check option (C): a = 1, b = 3, so sum = −(3)/1 = −3/1 = −3.
5. Check option (D): a = 3, b = 15, so sum = −(15)/3 = −15/3 = −5.
6. The equation with sum of roots −5 is option (D).
Answer: (D)
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Problem Set 2, Q1(7)
Choose the correct option for √5m² − √5m + √5 = 0: (A) real and unequal roots (B) real and equal roots (C) roots are not real (D) three roots
1. Compare the given equation √5m² − √5m + √5 = 0 with the standard form am² + bm + c = 0.
2. Identify the coefficients as a = √5, b = −√5, and c = √5.
3. Calculate the discriminant D using the formula D = b² − 4ac.
4. Substitute the values into the formula: D = (−√5)² − 4(√5)(√5).
5. Simplify the squares: D = 5 − 4(5).
6. Multiply the terms: D = 5 − 20.
7. Find the final value of the discriminant: D = −15.
8. Since D < 0, the roots are not real.
9. This corresponds to option (C).
Answer: (C)
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Problem Set 2, Q1(8)
Choose the correct option: one root of x² + mx − 5 = 0 is 2; m = (A) −2 (B) −1/2 (C) 1/2 (D) 2
1. Given equation is x² + mx − 5 = 0 and one root is 2.
2. Substitute x = 2 into the equation to find m.
3. 2² + m(2) − 5 = 0
4. 4 + 2m − 5 = 0
5. 2m − 1 = 0
6. 2m = 1
7. m = 1/2
8. The value of m is 1/2, which corresponds to option (C).
Answer: (C) 1/2
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Problem Set 2, Q2
Which of these are quadratic equations? (1) x² + 2x + 11 = 0 (2) x² − 2x + 5 = x² (3) (x + 2)² = 2x²
1. A quadratic equation must be of the form ax² + bx + c = 0, where a ≠ 0.
2. (1) The equation x² + 2x + 11 = 0 has a = 1, b = 2, c = 11.
3. (1) Since a = 1 ≠ 0, it is a quadratic equation.
4. (2) The equation x² − 2x + 5 = x² is simplified by subtracting x² from both sides.
5. (2) This gives −2x + 5 = 0, which is a linear equation.
6. (2) Since the x² term vanishes, it is not a quadratic equation.
7. (3) The equation (x + 2)² = 2x² is expanded to x² + 4x + 4 = 2x².
8. (3) Rearranging terms gives 0 = x² − 4x − 4.
9. (3) Since the coefficient of x² is 1 ≠ 0, it is a quadratic equation.
Answer: (1) quadratic; (2) not quadratic; (3) quadratic
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Problem Set 2, Q3
Find the discriminant: (1) 2y² − y + 2 = 0 (2) 5m² − m = 0 (3) √5x² − x − √5 = 0
1. (1) a = 2, b = −1, c = 2
2. (1) D = b² − 4ac
3. (1) D = (−1)² − 4 × 2 × 2
4. (1) D = 1 − 16
5. (1) D = −15
6. (2) a = 5, b = −1, c = 0
7. (2) D = b² − 4ac
8. (2) D = (−1)² − 4 × 5 × 0
9. (2) D = 1 − 0
10. (2) D = 1
11. (3) a = √5, b = −1, c = −√5
12. (3) D = b² − 4ac
13. (3) D = (−1)² − 4 × √5 × (−√5)
14. (3) D = 1 − (−20)
15. (3) D = 1 + 20
16. (3) D = 21
Answer: (1) -15; (2) 1; (3) 21
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Problem Set 2, Q4
One root of 2x² + kx − 2 = 0 is −2. Find k.
1. Given equation: 2x² + kx − 2 = 0
2. One root is x = −2
3. Substitute x = −2 into the equation: 2(−2)² + k(−2) − 2 = 0
4. Calculate the square: (−2)² = 4
5. Multiply by 2: 2 × 4 = 8
6. Equation becomes: 8 − 2k − 2 = 0
7. Combine constants: 8 − 2 = 6
8. Simplify: 6 − 2k = 0
9. Move 2k to the other side: 6 = 2k
10. Divide by 2: k = 6 ÷ 2
11. k = 3
Answer: k = 3
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Problem Set 2, Q5
Frame the quadratic equation with roots: (1) 10 and −10 (2) 1 − 3√5 and 1 + 3√5 (3) 0 and 7
1. (1) Sum of roots = 10 + (−10) = 0
2. (1) Product of roots = 10 × (−10) = −100
3. (1) Equation is x² − (sum)x + (product) = 0
4. (1) x² − 0x − 100 = 0
5. (1) x² − 100 = 0
6. (2) Sum of roots = (1 − 3√5) + (1 + 3√5) = 2
7. (2) Product of roots = (1 − 3√5)(1 + 3√5) = 1² − (3√5)²
8. (2) Product = 1 − 9 × 5 = 1 − 45 = −44
9. (2) Equation is x² − 2x + (−44) = 0
10. (2) x² − 2x − 44 = 0
11. (3) Sum of roots = 0 + 7 = 7
12. (3) Product of roots = 0 × 7 = 0
13. (3) Equation is x² − 7x + 0 = 0
14. (3) x² − 7x = 0
Answer: (1) x² - 100 = 0; (2) x² - 2x - 44 = 0; (3) x² - 7x = 0
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Problem Set 2, Q6
Find the nature of the roots: (1) 3x² − 5x + 7 = 0 (2) √3x² + √2x − 2√3 = 0 (3) m² − 2m + 1 = 0
1. (1) a = 3, b = -5, c = 7
2. (1) D = b² - 4ac
3. (1) D = (-5)² - 4(3)(7)
4. (1) D = 25 - 84
5. (1) D = -59
6. (1) Since D < 0, roots are not real
7. (2) a = √3, b = √2, c = -2√3
8. (2) D = b² - 4ac
9. (2) D = (√2)² - 4(√3)(-2√3)
10. (2) D = 2 - 4(-2)(3)
11. (2) D = 2 + 24
12. (2) D = 26
13. (2) Since D > 0, roots are real and unequal
14. (3) a = 1, b = -2, c = 1
15. (3) D = b² - 4ac
16. (3) D = (-2)² - 4(1)(1)
17. (3) D = 4 - 4
18. (3) D = 0
19. (3) Since D = 0, roots are real and equal
Answer: (1) not real; (2) real and unequal; (3) real and equal
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Problem Set 2, Q7(1)
Solve: 1/(x + 5) = 1/x².
1. Given equation: 1/(x + 5) = 1/x²
2. Cross-multiply to get x² = x + 5
3. Rearrange into standard form: x² − x − 5 = 0
4. Identify coefficients: a = 1, b = −1, c = −5
5. Calculate discriminant D = b² − 4ac = (−1)² − 4(1)(−5)
6. D = 1 + 20 = 21
7. Apply quadratic formula: x = (−b ± √D) / 2a
8. x = (1 ± √21) / 2
9. The solutions are x = (1 + √21)/2 and x = (1 − √21)/2
Answer: x = (1 + √21)/2, x = (1 − √21)/2
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Problem Set 2, Q7(2)
Solve: x² − 3x/10 − 1/10 = 0.
1. Multiply the equation by 10 to clear the denominators: 10x² − 3x − 1 = 0
2. Split the middle term −3x into −5x + 2x so the factors match: 10x² − 5x + 2x − 1 = 0
3. Group the terms: (10x² − 5x) + (2x − 1) = 0
4. Factor out the common terms from each group: 5x(2x − 1) + 1(2x − 1) = 0
5. Factor out the common binomial (2x − 1): (2x − 1)(5x + 1) = 0
6. Set each factor to zero: 2x − 1 = 0 or 5x + 1 = 0
7. Solve the first equation: 2x = 1, so x = 1/2
8. Solve the second equation: 5x = −1, so x = −1/5
Answer: x = 1/2 or x = −1/5
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Problem Set 2, Q7(3)
Solve: (2x + 3)² = 25.
1. Given (2x + 3)² = 25
2. Taking square root on both sides, 2x + 3 = ±5
3. Case 1: 2x + 3 = 5
4. 2x = 5 − 3
5. 2x = 2
6. x = 2 ÷ 2
7. x = 1
8. Case 2: 2x + 3 = −5
9. 2x = −5 − 3
10. 2x = −8
11. x = −8 ÷ 2
12. x = −4
Answer: x = 1 or x = −4
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Problem Set 2, Q7(4)
Solve: m² + 5m + 5 = 0.
1. Identify the coefficients of the quadratic equation m² + 5m + 5 = 0 as a = 1, b = 5, and c = 5.
2. Write the quadratic formula for the roots: m = (−b ± √(b² − 4ac)) / 2a.
3. Substitute the values into the formula: m = (−5 ± √(5² − 4 × 1 × 5)) / (2 × 1).
4. Simplify the expression inside the square root: 5² − 4 × 1 × 5 = 25 − 20 = 5.
5. Write the simplified formula: m = (−5 ± √5) / 2.
6. The two roots are m = (−5 + √5) / 2 and m = (−5 − √5) / 2.
Answer: m = (-5 + √5)/2, m = (-5 - √5)/2
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Problem Set 2, Q7(5)
Solve: 5m² + 2m + 1 = 0.
1. Given equation: 5m² + 2m + 1 = 0
2. Compare with am² + bm + c = 0, so a = 5, b = 2, c = 1
3. Calculate discriminant D = b² − 4ac
4. D = 2² − 4 × 5 × 1
5. D = 4 − 20
6. D = −16
7. Since D < 0, the roots are not real
Answer: not real
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Problem Set 2, Q7(6)
Solve: x² − 4x − 3 = 0.
1. Given equation is x² − 4x − 3 = 0
2. Here a = 1, b = −4, c = −3
3. D = b² − 4ac = (−4)² − 4(1)(−3)
4. D = 16 + 12
5. D = 28
6. Using the formula x = (−b ± √D) / 2a
7. x = (4 ± √28) / 2
8. Since √28 = √(4 × 7) = 2√7, substitute this value
9. x = (4 ± 2√7) / 2
10. Divide numerator and denominator by 2
11. x = 2 ± √7
12. Thus, x = 2 + √7 or x = 2 − √7
Answer: x = 2 + √7 or x = 2 − √7
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Problem Set 2, Q8
Find m if (m − 12)x² + 2(m − 12)x + 2 = 0 has real and equal roots.
1. The given quadratic equation is (m − 12)x² + 2(m − 12)x + 2 = 0.
2. Here, a = m − 12, b = 2(m − 12), and c = 2.
3. For real and equal roots, the discriminant D must be zero.
4. The formula for the discriminant is D = b² − 4ac.
5. Substitute the values into the formula: D = [2(m − 12)]² − 4(m − 12)(2).
6. Simplify the expression: D = 4(m − 12)² − 8(m − 12).
7. Factor out the common term 4(m − 12): D = 4(m − 12)[(m − 12) − 2].
8. Simplify the term inside the bracket: D = 4(m − 12)(m − 14).
9. Set the discriminant to zero: 4(m − 12)(m − 14) = 0.
10. This gives two possible cases: m − 12 = 0 or m − 14 = 0.
11. If m − 12 = 0, then m = 12. However, if m = 12, the coefficient of x² becomes 0, so it is not a quadratic equation.
12. Therefore, we must have m − 14 = 0.
13. Solving for m gives m = 14.
Answer: m = 14
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Problem Set 2, Q9
The roots of a quadratic equation add up to 5 and their cubes add up to 35. Find the equation.
1. Let the roots of the quadratic equation be α and β.
2. Given that the sum of the roots is 5, so α + β = 5.
3. Given that the sum of the cubes of the roots is 35, so α³ + β³ = 35.
4. Use the identity α³ + β³ = (α + β)³ − 3αβ(α + β).
5. Substitute the known values into the identity: 35 = (5)³ − 3αβ(5).
6. Calculate the cube: 35 = 125 − 15αβ.
7. Rearrange to find the product: 15αβ = 125 − 35.
8. Simplify: 15αβ = 90.
9. Solve for the product: αβ = 90 ÷ 15 = 6.
10. The quadratic equation is x² − (sum of roots)x + (product of roots) = 0.
11. Substitute the values: x² − 5x + 6 = 0.
Answer: x² - 5x + 6 = 0
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Problem Set 2, Q11
Mukund has ₹50 more than Sagar, and the product of their amounts is 15,000. Find how much each has.
1. Let the amount with Sagar be ₹x.
2. Then the amount with Mukund is ₹(x + 50).
3. Product of their amounts = 15000.
4. x(x + 50) = 15000.
5. x² + 50x − 15000 = 0.
6. x² + 150x − 100x − 15000 = 0
7. x(x + 150) − 100(x + 150) = 0
8. (x + 150)(x − 100) = 0
9. x = −150 or x = 100.
10. Amount cannot be negative, so x = 100.
11. Sagar has ₹100.
12. Mukund has ₹(100 + 50) = ₹150.
Answer: Sagar ₹100; Mukund ₹150
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Problem Set 2, Q12
The squares of two numbers differ by 120, and the square of the smaller number is twice the greater number. Find the numbers.
1. Let the greater number be x and the smaller number be y.
2. From the second condition, y² = 2x, so x = y²/2.
3. From the first condition, x² − y² = 120.
4. Substitute x = y²/2 into the equation: (y²/2)² − y² = 120.
5. Simplify to get y⁴/4 − y² = 120.
6. Multiply by 4 to clear the denominator: y⁴ − 4y² = 480.
7. Rearrange as y⁴ − 4y² − 480 = 0.
8. Let z = y², so z² − 4z − 480 = 0.
9. Factor the quadratic: (z − 24)(z + 20) = 0.
10. Since z = y² must be positive, z = 24.
11. Thus y² = 24, so y = √24 or y = −√24.
12. Calculate x using x = y²/2: x = 24/2 = 12.
Answer: 12 and √24, or 12 and −√24
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Problem Set 2, Q13
540 oranges are shared equally among some students. With 30 more students each would get 3 oranges fewer. Find the number of students.
1. Let the number of students be x.
2. Oranges per student = 540/x.
3. With 30 more students, oranges per student = 540/(x + 30).
4. Difference in oranges per student is 3, so 540/x − 540/(x + 30) = 3.
5. Combine fractions: [540(x + 30) − 540x] / [x(x + 30)] = 3.
6. Simplify numerator: 540x + 16200 − 540x = 16200.
7. Equation becomes 16200 / [x(x + 30)] = 3.
8. Multiply both sides by x(x + 30): 16200 = 3x(x + 30).
9. Divide by 3: 5400 = x² + 30x.
10. Rearrange to standard form: x² + 30x − 5400 = 0.
11. Factorize: (x + 90)(x − 60) = 0.
12. Solutions are x = −90 or x = 60.
13. Number of students cannot be negative, so x = 60.
Answer: 60
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Problem Set 2, Q14
A farm's length is 10 m more than twice its breadth. A square pond inside has side one third of the farm's breadth, and the farm's area is 20 times the pond's area. Find the farm's length and breadth and the pond's side.
1. Let the breadth of the farm be b m.
2. The length is 10 m more than twice the breadth, so length = 2b + 10 m.
3. The side of the square pond is one third of the breadth, so side = b/3 m.
4. Area of the farm = length × breadth = (2b + 10)b = 2b² + 10b.
5. Area of the pond = side² = (b/3)² = b²/9.
6. The farm's area is 20 times the pond's area, so 2b² + 10b = 20(b²/9).
7. Multiply both sides by 9 to clear the denominator: 18b² + 90b = 20b².
8. Rearrange the equation: 20b² − 18b² − 90b = 0.
9. Simplify: 2b² − 90b = 0.
10. Factor out 2b: 2b(b − 45) = 0.
11. So b = 0 or b = 45.
12. Since breadth cannot be zero, b = 45 m.
13. Length = 2(45) + 10 = 90 + 10 = 100 m.
14. Side of pond = 45/3 = 15 m.
Answer: breadth 45 m; length 100 m; side of the pond 15 m
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Problem Set 2, Q15
With both taps open a tank fills in 2 hours. Alone, the smaller tap takes 3 hours more than the larger. How long does each tap take alone?
1. Let the time taken by the larger tap alone be x hours.
2. Then the time taken by the smaller tap alone is (x + 3) hours.
3. The rate of the larger tap is 1/x and the rate of the smaller tap is 1/(x + 3).
4. Since both taps together fill the tank in 2 hours, their combined rate is 1/2.
5. Equation: 1/x + 1/(x + 3) = 1/2.
6. Combine fractions: (x + 3 + x) / (x(x + 3)) = 1/2.
7. Simplify numerator: (2x + 3) / (x² + 3x) = 1/2.
8. Cross-multiply: 2(2x + 3) = x² + 3x.
9. Expand left side: 4x + 6 = x² + 3x.
10. Rearrange to form a quadratic equation: x² + 3x - 4x - 6 = 0.
11. Simplify: x² - x - 6 = 0.
12. Factorize: (x - 3)(x + 2) = 0.
13. Solutions: x = 3 or x = -2.
14. Time cannot be negative, so x = 3.
15. Larger tap takes 3 hours.
16. Smaller tap takes x + 3 = 3 + 3 = 6 hours.
Answer: larger tap 3 hours, smaller tap 6 hours
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Other chapters solved for this book

Chapter 1: Linear Equations in Two Variables — 58 questions solved
Chapter 3: Arithmetic Progression — 47 questions solved
Chapter 5: Probability — 25 questions solved

← All state-board textbook solutions

Maharashtra SSC Class 10 Mathematics Chapter 2 Quadratic Equations: textbook solutions

Practice Set 2.1

Practice Set 2.1, Q2

Practice Set 2.1, Q3

Practice Set 2.1, Q4

Practice Set 2.1, Q5

Practice Set 2.1, Q6

Practice Set 2.2

Practice Set 2.2, Q1(1)

Practice Set 2.2, Q1(2)

Practice Set 2.2, Q1(3)

Practice Set 2.2, Q1(4)

Practice Set 2.2, Q1(5)

Practice Set 2.2, Q1(6)

Practice Set 2.2, Q1(8)

Practice Set 2.2, Q1(9)

Practice Set 2.2, Q1(10)

Practice Set 2.2, Q1(11)

Practice Set 2.2, Q1(12)

Practice Set 2.3

Practice Set 2.3, Q1(1)

Practice Set 2.3, Q1(2)

Practice Set 2.3, Q1(3)

Practice Set 2.3, Q1(4)

Practice Set 2.3, Q1(5)

Practice Set 2.3, Q1(6)

Practice Set 2.4

Practice Set 2.4, Q1

Practice Set 2.4, Q2(1)

Practice Set 2.4, Q2(2)

Practice Set 2.4, Q2(3)

Practice Set 2.4, Q2(4)

Practice Set 2.4, Q2(5)

Practice Set 2.4, Q2(6)

Practice Set 2.4, Q3

Practice Set 2.5

Practice Set 2.5, Q1

Practice Set 2.5, Q2

Practice Set 2.5, Q3

Practice Set 2.5, Q4

Practice Set 2.5, Q5

Practice Set 2.5, Q6

Practice Set 2.5, Q7

Practice Set 2.6

Practice Set 2.6, Q1

Practice Set 2.6, Q2

Practice Set 2.6, Q3

Practice Set 2.6, Q4

Practice Set 2.6, Q5

Practice Set 2.6, Q6

Practice Set 2.6, Q7

Practice Set 2.6, Q8

Practice Set 2.6, Q9

Practice Set 2.6, Q10

Problem Set 2

Problem Set 2, Q1(1)

Problem Set 2, Q1(2)

Problem Set 2, Q1(3)

Problem Set 2, Q1(4)

Problem Set 2, Q1(5)

Problem Set 2, Q1(6)

Problem Set 2, Q1(7)

Problem Set 2, Q1(8)

Problem Set 2, Q2

Problem Set 2, Q3

Problem Set 2, Q4

Problem Set 2, Q5

Problem Set 2, Q6

Problem Set 2, Q7(1)

Problem Set 2, Q7(2)

Problem Set 2, Q7(3)

Problem Set 2, Q7(4)

Problem Set 2, Q7(5)

Problem Set 2, Q7(6)

Problem Set 2, Q8

Problem Set 2, Q9

Problem Set 2, Q11

Problem Set 2, Q12

Problem Set 2, Q13

Problem Set 2, Q14