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NCERT Class 10 Mathematics Exercise 4.1 Solutions: Quadratic Equations

Step-by-step working for every question of Exercise 4.1 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 4.1 Q1

   Which are quadratic equations? (i) (x + 1)² = 2(x − 3) (ii) x² − 2x = (−2)(3 − x) (iii) (x − 2)(x + 1) = (x − 1)(x + 3) (iv) (x − 3)(2x + 1) = x(x + 5) (v) (2x − 1)(x − 3) = (x + 5)(x − 1) (vi) x² + 3x + 1 = (x − 2)² (vii) (x + 2)³ = 2x(x² − 1) (viii) x³ − 4x² − x + 1 = (x − 2)³.

   1. (i) Expand both sides to get x² + 2x + 1 = 2x − 6, which simplifies to x² + 7 = 0, so it is a quadratic equation.
      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.

   2. (ii) Expand the right side to get x² − 2x = −6 + 2x, which simplifies to x² − 4x + 6 = 0, so it is a quadratic equation.
      why?

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

   3. (iii) Expand both sides to get x² − x − 2 = x² + 2x − 3, which simplifies to −3x + 1 = 0, so it is not a quadratic equation.
      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.

   4. (iv) Expand both sides to get 2x² − 5x − 3 = x² + 5x, which simplifies to x² − 10x − 3 = 0, so it is a quadratic equation.
      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.

   5. (v) Expand both sides to get 2x² − 7x + 3 = x² + 4x − 5, which simplifies to x² − 11x + 8 = 0, so it is a quadratic equation.
      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. (vi) Expand the right side to get x² + 3x + 1 = x² − 4x + 4, which simplifies to 7x − 3 = 0, so it is not a quadratic equation.
      why?

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

   7. (vii) Expand the left side to get x³ + 6x² + 12x + 8 = 2x³ − 2x, which simplifies to x³ − 6x² − 14x − 8 = 0, so it is not a quadratic equation.
      why?

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

   8. (viii) Expand the right side to get x³ − 4x² − x + 1 = x³ − 6x² + 12x − 8, which simplifies to 2x² − 13x + 9 = 0, so it is a quadratic equation.
      why?

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

   Answer: (i) yes; (ii) yes; (iii) no; (iv) yes; (v) yes; (vi) no; (vii) no; (viii) yes

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2. Ex 4.1 Q2

   Write as quadratic equations: (i) a rectangle of area 528 m² whose length is 1 more than twice its breadth (ii) two consecutive positive integers with product 306 (iii) Rohan's mother is 26 years older; in 3 years the product of their ages will be 360 (iv) a train covers 480 km; 8 km/h slower it would take 3 hours more.

   1. (i) Let the breadth be x m, so the length is (2x + 1) m. The area is length times breadth, giving x(2x + 1) = 528. Expanding and rearranging gives 2x² + x − 528 = 0.
      why?

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

   2. (ii) Let the first integer be x, so the next is (x + 1). Their product is x(x + 1) = 306. Expanding and rearranging gives x² + x − 306 = 0.
      why?

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

   3. (iii) Let Rohan's age be x, so his mother's age is (x + 26). In 3 years, their ages are (x + 3) and (x + 29). The product is (x + 3)(x + 29) = 360. Expanding gives x² + 32x + 87 = 360. Rearranging gives x² + 32x − 273 = 0.
      why?

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

   4. (iv) Let the speed be x km/h. The time taken is 480/x hours. If the speed is (x − 8) km/h, the time is 480/(x − 8) hours. The difference is 3 hours, so 480/(x − 8) − 480/x = 3. Simplifying the left side gives 3840/(x(x − 8)) = 3. Multiplying by x(x − 8) gives 3840 = 3x² − 24x. Dividing by 3 gives 1280 = x² − 8x. Rearranging gives x² − 8x − 1280 = 0.
      why?

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

   Answer: (i) 2x² + x - 528 = 0; (ii) x² + x - 306 = 0; (iii) x² + 32x - 273 = 0; (iv) x² - 8x - 1280 = 0

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