NCERT Solutions For Class 10 Maths Chapter 2 Polynomials

 NCERT Solutions For Class 10 Maths Chapter 2 Polynomials

Exercise 2.1

1. The graphs of `y = p(x)` are given in Fig. below, for some polynomials `p(x)`. Find the number of zeroes of `p(x)`, in each case.

Solutions:

`(i)` No zeroes   `(ii) 1  (iii) 3  (iv) 2  (v) 4  (vi) 3` 

Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 

`(i)  x^2 – 2x – 8   (ii) 4s^2 – 4s + 1  (iii) 6x^2 – 3 – 7x`  

`(iv)  4u^2 + 8u   (v)  t^2 – 15   (vi) 3x^2 – x – 4`

Solutions

`(i)  x^2 – 2x – 8`

  `x^2 – 2x – 8 = 0`

`⇒ x^2 -4x + 2x -8 = 0`

`⇒ x(x + 2) -4 (x + 2) = 0`

`⇒ (x + 2) (x -4) = 0`

`⇒ x = -2`  or   `x = 4` 

`alpha + beta = -b/a`

`⇒ -2 + 4 = 2/1`

`alpha beta = c/a`

`⇒ -2 × 4 = -8/1`

verified.

Video Solution (Click here) 🎥

`(ii)  4s^2 – 4s + 1`

`⇒ 4s^2 -2s -2s + 1 = 0`

`⇒ 2s(2s -1) - 1(2s -1) = 0`

`⇒ (2s -1) (2s -1) = 0`

`⇒ s = 1/2`  or   `s = 1/2`

`alpha + beta = -b/a`

`⇒ 1/2 + 1/2 = 2/2 = 1 = 4/4`

`alpha beta = c/a`

`⇒ 1/2 × 1/2 = 1/4`

verified.

`(iii)  6x^2 – 3 – 7x`

  `6x^2 – 7x – 3 = 0`

`⇒ 6x^2 -9x + 2x -3 = 0`

`⇒ 3x(2x -3) +1 (2x -3) = 0`

`⇒ (2x -3) (3x +1) = 0`

`⇒ x = 3/2`  or   `x = -1/3` 

`alpha + beta = -b/a`

`⇒ 3/2 + (-1/3 )=  3/2 -1/3 = frac{9-2}{6} = 7/6`

`alpha beta = c/a`

`⇒ 3/2 × (-1/3) = - 1/2 = -3/6`

verified.

Video Solutions (ii), (iii) (Click here ) 🎥

`(iv)  4u^2 + 8u`

`⇒ 4u (u + 2) = 0`

`⇒ u = 0` or `u = -2`

`alpha + beta = -b/a`

`⇒ 0 + (-2) = -2 = -8/4 `

`alpha beta = c/a`

`⇒ 0 × (-2) = 0`

verified.

`(v)  t^2 – 15`

`t^2 – (sqrt15)^2 = 0`

`⇒ (t + sqrt15)(t -sqrt15) = 0`

`⇒ t = sqrt15`  or  `t = -sqrt15`

`alpha + beta = -b/a`

`⇒ sqrt15 + (-sqrt15) = 0`

`alpha beta = c/a`

`⇒ sqrt15 × (-sqrt15) = -15/1`

verified.

Video Solutions (iv), (v) (Click here) 🎥

`(vi)  3x^2 – x – 4`

  `3x^2 – x – 4 = 0`

`⇒ 3x^2 -4x + 3x -4 = 0`

`⇒ x(3x -4) +1 (3x -4) = 0`

`⇒ (3x -4) (x +1) = 0`

`⇒ x = 4/3`  or   `x = -1` 

`alpha + beta = -b/a`

`⇒ 4/3 + (-1) = 4/3 -1 = frac{4-3}{3} = 1/3`

`alpha beta = c/a`

`⇒ 4/3 × (-1) = -4/3`

verified.

Video Solution (vi) 🎥

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

`(i) 1/4, -1    (ii) sqrt2, 1/3      (iii)  0, sqrt5   (iv)  1,1   (v)  -1/4, 1/4   (vi)  4,1`

Solutions

`(i) 1/4, -1` 

`alpha + beta = 1/4`

`alpha beta = -1`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - 1/4x -1`

`= 4x^2 -x -4`

 `(ii) sqrt2, 1/3`  

`alpha + beta = sqrt2`

`alpha beta = 1/3`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - sqrt2x + 1/3`

`= 3x^2 -3sqrt2x +1`

  `(iii)  0, sqrt5`  

`alpha + beta = 0`

`alpha beta = sqrt5`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - 0x + sqrt5`

`= x^2 + sqrt5`

`(iv)  1,1`

 `alpha + beta = 1`

`alpha beta = 1`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - 1x + 1`

`= x^2 -x +  1`

`(v)  -1/4, 1/4`

 `alpha + beta = -1/4`

`alpha beta = 1/4`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - (-1/4)x + 1/4`

`= x^2 +1/4x +  1/4`

`= 4x^2 +x +  1`

 `(vi)  4,1`

 `alpha + beta = 4`

`alpha beta = 1`

`therefore` required quadratic polynomial is 

`x^2 - (alpha + beta)x + alpha beta`

`= x^2 - 4x + 1`