# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

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 Book Name Class 10 Mathematics NCERT Textbook Chapter Chapter 2 Polynomials Exercise Ex 2.2

# NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Exercise 2.2

1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8
(ii) 4s2 – 4s + 1
(iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u
(v) t2 – 15
(vi) 3x2 – x – 4

Solution

(i) x2 – 2x – 8 = x2 -4x +4x -8

= x(x-) + 2(x-4) = (x+2)(x-4)

So, the value of p(x) = x2 – 2x – 8 is zero when (x+2)=0 and (x-4)=0

When x = -2 or x = . This, the zeroes of x2 – 2x – 8 are 4 and -2.

∴ Sum of the zeroes = (-2) + (4) =2 = -(-2)/1 = -(coefficient of x)/coefficient of x2

Product of the zeroes = (-2) × (4) = -8 = (-8)/1 = constant term / coefficient of x2

(ii) 4s2 – 4s + 1 = 4s2 - 2s - 2s + 1

= 2s(2s -1) - 1(2s-1)

= (2s -1) (2s-1)

So, the value of 4s2 -2s +1 is zero when either (2s-1)=0

or (2s -1) = 0, when x = 1/2 or x =1/2.

Thus, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.

∴ Sum of the zeroes = 1/2 + 1/2 = 1 = -(-1)/1 = -(coefficient of s)/ coefficient of s2

and, product of zeroes = (1/2) (1/2) = 1/4 = Constant term/Coefficient of s2

(iii) 6x2 – 3 – 7x = 6x2 6x– 7– 3

6x– 9– 2x -3

= 3x(2x -3) +1(2x-3)

= (3x + 1) (2x - 3)

The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.

(iv) 4u2 + 8u = 4u2 + 8u + 0
= 4u(u + 2)

The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u2 + 8u are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u2.

(v) t2 – 15
= t- 0.t - 15
= (t - √15) (t + √15)
The value of t2 - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t2 - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u2.

(vi) 3x2 – x – 4
= (3x - 4) (x + 1)
The value of 3x2 – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 – x – 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x2.

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

(i) 1/4 , -1

(ii) √2 , 1/3

(iii) 0, √5

(iv) 1,1

(v) -1/4 ,1/4

(vi) 4,1

Solution

(i) 1/4 , -1

Let the quadratic polynomial be ax2 + bx + c and its zeroes α and β.

α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 - x -4.

(ii) √2 , 1/3

Let the quadratic polynomial be ax2 + bx + c and its zeroes α and β.

α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x2 -3√2x +1.

(iii) 0, √5

Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a
αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x2 + √5.

(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.

(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x +1.

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