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NCERT Class 10 Mathematics Exercise 4.2 Solutions: Quadratic Equations

Step-by-step working for every question of Exercise 4.2 in NCERT's Class 10 Mathematics book, chapter “Quadratic Equations”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Ex 4.2 Q1

   Find the roots by factorisation: (i) x² − 3x − 10 = 0 (ii) 2x² + x − 6 = 0 (iii) √2x² + 7x + 5√2 = 0 (iv) 2x² − x + 1/8 = 0 (v) 100x² − 20x + 1 = 0.

   1. (i) x² − 3x − 10 = x² − 5x + 2x − 10 = x(x − 5) + 2(x − 5) = (x − 5)(x + 2) = 0, so x = 5 or x = −2.
   2. (ii) 2x² + x − 6 = 2x² + 4x − 3x − 6 = 2x(x + 2) − 3(x + 2) = (2x − 3)(x + 2) = 0, so x = 3/2 or x = −2.
   3. (iii) The product a × c = √2 × 5√2 = 10, so split 7x as 5x + 2x: √2x² + 5x + 2x + 5√2 = 0.
   4. (iii) Group: x(√2x + 5) + √2(√2x + 5) = (√2x + 5)(x + √2) = 0, so x = −5/√2 or x = −√2.
   5. (iv) Multiply by 8: 16x² − 8x + 1 = 0, which is (4x − 1)² = 0, so x = 1/4 (a repeated root).
   6. (v) 100x² − 20x + 1 = (10x − 1)² = 0, so x = 1/10 (a repeated root).

   Answer: (i) −2, 5; (ii) −2, 3/2; (iii) −5/√2, −√2; (iv) 1/4, 1/4; (v) 1/10, 1/10

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2. Ex 4.2 Q2

   Solve the two situations of the chapter's Example 1: (i) John and Jivanti have 45 marbles; after each loses 5 the product is 124 (ii) toys cost (55 − number made) rupees each and the day's total cost was ₹750.

   1. (i) Let the number of marbles John has be x, so Jivanti has 45 − x.
   2. (i) After losing 5 each, their counts are x − 5 and 40 − x.
   3. (i) The product is 124, so (x − 5)(40 − x) = 124.
   4. (i) Expanding gives 40x − x² − 200 + 5x = 124, which simplifies to x² − 45x + 324 = 0.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   5. (i) Factoring the quadratic yields (x − 36)(x − 9) = 0, so x = 36 or x = 9.
   6. (i) If John has 36, Jivanti has 9; if John has 9, Jivanti has 36.
   7. (ii) Let the number of toys made be x, so the cost per toy is 55 − x.
   8. (ii) The total cost is 750, so x(55 − x) = 750.
   9. (ii) This simplifies to 55x − x² = 750, or x² − 55x + 750 = 0.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   10. (ii) Factoring gives (x − 25)(x − 30) = 0, so x = 25 or x = 30.

   Answer: (i) 36 and 9; (ii) 25 or 30

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3. Ex 4.2 Q3

   Find two numbers with sum 27 and product 182.

   1. Let the two numbers be x and 27 − x.
   2. Their product is given as 182, so x(27 − x) = 182.
   3. Expanding gives 27x − x² = 182.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   4. Rearranging into standard form yields x² − 27x + 182 = 0.
   5. Factorising the quadratic: x² − 13x − 14x + 182 = 0.
   6. Grouping terms: x(x − 13) − 14(x − 13) = 0.
   7. Taking out the common factor: (x − 13)(x − 14) = 0.
   8. So x = 13 or x = 14.
   9. If x = 13, the other number is 27 − 13 = 14.
   10. If x = 14, the other number is 27 − 14 = 13.

   Answer: 13 and 14

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4. Ex 4.2 Q4

   Find two consecutive positive integers whose squares add up to 365.

   1. Let the two consecutive positive integers be x and x + 1.
   2. According to the question, the sum of their squares is 365, so x² + (x + 1)² = 365.
   3. Expanding the equation gives x² + x² + 2x + 1 = 365.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   4. Simplifying the terms results in 2x² + 2x + 1 = 365.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   5. Subtracting 365 from both sides gives 2x² + 2x − 364 = 0.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   6. Dividing the entire equation by 2 yields x² + x − 182 = 0.
   7. Factoring the quadratic equation by splitting the middle term gives x² + 14x − 13x − 182 = 0.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   8. Grouping terms leads to x(x + 14) − 13(x + 14) = 0, which factors to (x + 14)(x − 13) = 0.
   9. Solving for x gives x = −14 or x = 13.
   10. Since the integers must be positive, we take x = 13, making the other integer x + 1 = 14.

   Answer: 13 and 14

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5. Ex 4.2 Q5

   A right triangle's altitude is 7 cm less than its base and the hypotenuse is 13 cm. Find the other two sides.

   1. Let the base of the right triangle be x cm.
   2. The altitude is 7 cm less than the base, so it is (x − 7) cm.
   3. The hypotenuse is given as 13 cm.
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   4. By the Pythagoras theorem, base² + altitude² = hypotenuse².
      why?

      In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

   5. Substitute the values: x² + (x − 7)² = 13².
      why?

      Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.

   6. Expand the terms: x² + x² − 14x + 49 = 169.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   7. Combine like terms and simplify: 2x² − 14x − 120 = 0.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   8. Divide the entire equation by 2: x² − 7x − 60 = 0.
   9. Factorize the quadratic equation: (x − 12)(x + 5) = 0.
   10. Solve for x: x = 12 or x = −5.
   11. Since length cannot be negative, the base is 12 cm.
   12. Calculate the altitude: 12 − 7 = 5 cm.

   Answer: base 12 cm, altitude 5 cm

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6. Ex 4.2 Q6

   Each pottery article costs ₹3 more than twice the number made that day, and the day's total cost was ₹90. Find the number of articles and the cost of each.

   1. Let the number of articles made be x.
   2. The cost of each article is given as ₹3 more than twice the number made, so cost = 2x + 3.
   3. The total cost for the day is the product of the number of articles and the cost per article.
   4. Set up the equation: x(2x + 3) = 90.
   5. Expand the equation to get 2x² + 3x = 90.
      why?

      Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

   6. Rearrange into standard form: 2x² + 3x − 90 = 0.
   7. Factorize by splitting the middle term: 2x² + 15x − 12x − 90 = 0.
      why?

      Splitting the middle term into two parts whose product equals (first coefficient × constant) lets the expression be grouped into two brackets; a product is zero only when one of its factors is zero.

   8. Group terms: x(2x + 15) − 6(2x + 15) = 0.
   9. Factor out the common binomial: (x − 6)(2x + 15) = 0.
   10. Solve for x: x = 6 or x = −15/2.
   11. Since the number of articles cannot be negative, x = 6.
   12. Calculate the cost per article: 2(6) + 3 = 12 + 3 = 15.

   Answer: 6 articles, ₹15 each

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