NCERT Solutions for Class 10 Maths Ex 2.2

# NCERT Solutions for Class 10 Maths Ex 2.2

In this post, you will find the NCERT solutions for class 10 maths ex 2.2. These solutions are based on the latest curriculum of NCERT maths class 10.

## NCERT Solutions for Class 10 Maths Ex 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

We have, x2 2x 8 = x2 − 4x + 2x − 8

x(x − 4) + 2(x − 4) = (x − 4)( x + 2)

To find the zeroes of this polynomial, equate the given equation to 0.

x2 2x 8 = 0

(x − 4)(x + 2) = 0

= 4, −2

Therefore, two zeroes of this polynomial are 4 and −2.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 1, b = 2 and c = 8

Sum of zeroes = 4 + (– 2) = 2 = −(−2)/1

= −b/a = − Coefficient of x/Coefficient of x2

Product of zeroes = 4 × −2 = −8

= −8/1 = c/a = Constant term/Coefficient of x2

(ii) 4s2 4s + 1

We have, 4s2 4s + 1 = 4s2 − 2s − 2s + 1

=2s(2s−1)−1(2s−1)
= (2s−1)(2s−1)

To find the zeroes of this polynomial, equate the given equation to 0.

4s2 4s + 1 = 0

(2s − 1)(2s − 1) = 0

= 1/2, 1/2

Therefore, two zeroes of this polynomial are 1/2 and 1/2.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 4, b = -4 and c = 1

Sum of zeroes = 1/2 + 1/2 = 1 = −(−1)/1 × 4/4 = −(−4)/4

= −b/a = −Coefficient of x/Coefficient of x2

Product of zeroes = 1/2 × 1/2 = 1/4

= c/a = Constant term/Coefficient of x2

(iii) 6x2 – 3 − 7x

We have, 6x2 – 3 − 7x = 6x2 − 7x − 3

= 6x2 − 9x + 2x − 3

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

To find the zeroes of this polynomial, equate the given equation to 0.

6x2 – 3 − 7x = 0

(2x − 3)(3x + 1) = 0

= 3/2, −1/3

Therefore, two zeroes of this polynomial are 3/2 and −1/3.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 6, b = −7 and c = −3

Sum of zeroes = 3/2 + −1/3 = (9−2)/6

= 7/6 = −(−7)/6

= −b/a = −Coefficient of x/Coefficient of x2

Product of zeroes = 3/2 × −1/3

= −1/2 = −3/6

= c/a = Constant term/Coefficient of x2

(iv) 4u2 + 8u

We have, 4u2 + 8u = 4u(u + 2)

To find the zeroes of this polynomial, equate the given equation to 0.

4u2 + 8u = 0

4u(u + 2) = 0

= 0, −2

Therefore, two zeroes of this polynomial are 0 and −2.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 4, b = 8 and c = 0

Sum of zeroes = 0−2 = −2

= −2/1 × 4/4 = −8/4

= −b/a = −Coefficient of x/Coefficient of x2

Product of zeroes = 0 × −2 = 0

= 0/4 = c/a = Constant term/Coefficient of x2

(v) t2 15

We have, t2 – 15 = 0

t2 = 15

= ±√15

Therefore, two zeroes of this polynomial are √15 and −√15.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 1, b = 0 and c = −15

Sum of zeroes = √15 + (−15) = 0 = 0/1

= −b/a = −Coefficient of x/Coefficient of x2

Product of Zeroes = √15 × (−√15) = −15 = −15/1

= c/a = Constant term/Coefficient of x2

(vi) 3x2 – x − 4

We have, 3x2 – x − 4  = 3x2 − 4x + 3x − 4

x(3x − 4) + 1(3x − 4)

= (3x − 4)(x +  1)

To find the zeroes of this polynomial, equate the given equation to 0.

3x2 – x − 4 = 0

(3x − 4)(x + 1) = 0

= 4/3, −1

Therefore, two zeroes of this polynomial are  4/3 and −1.

Comparing given polynomial with general form ax2 + bx + c, we get

a = 3, b = −1 and c = −4

Sum of zeroes = 4/3 + (−1) = (4−3)/3

= 1/3 = −(−1)/3

=  −b/a = −Coefficient of x/Coefficient of x2

Product of zeroes = 4/3 × (−1) = −4/3

= c/a = 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, 13

(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.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = 1/4 = −b/a

And product of zeroes = Î± × Î² = −1 = −1/1 × 4/4 = −4/4 = c/a

Thus, a = 4, b = −1, c = −4

Therefore, the quadratic polynomial which satisfies the above conditions is 4x2 – x – 4.

(ii) √2, 1/3

Let the quadratic polynomial be ax2 + bx + c.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = √2 = 3√2/3 = −b/a

And product of zeroes = Î± × Î² = 1/3 = 1/3 = c/a

Thus, a = 3, b = −3√2, c =1

Therefore, quadratic polynomial which satisfies the above conditions is 3x2 − 3√2x + 1.

(iii) 0, √5

Let the quadratic polynomial be ax2 + bx + c.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = 0 = 0/1 = −b/a

And product of zeroes = Î± × Î² = √5 = √5/1 = c/a

Thus, a = 1, b = 0, c = √5

Therefore, the quadratic polynomial which satisfies the above conditions is x2 + √5.

(iv) 1, 1

Let the quadratic polynomial be ax2 + bx + c.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = 1 = −(−1)/1 = −b/a

And product of zeroes = Î± × Î² = 1 = 1/1 = c/a

Thus, a = 1, b = −1, c = 1

Therefore, the quadratic polynomial which satisfies the above conditions is x2 – x + 1.

(v) −1/4, 1/4

Let the quadratic polynomial be ax2 + bx + c.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = −1/4 = −b/a

And product of zeroes = Î± × Î² = 1/4 = c/a

Thus, a = 4 , b = 1 , c = 1

Therefore, the quadratic polynomial which satisfies the above conditions is 4x2 + x + 1.

(vi) 4, 1

Let the quadratic polynomial be ax2 + bx + c.

Let Î± and Î² be two zeroes of above quadratic polynomial.

Then, sum of zeroes = Î± + Î² = 4  = −(−4)/1 = −b/a

And product of zeroes = Î± × Î² = 1 = 1/1 = c/a

Thus, a = 1 , b = −4 , c = 1

Therefore, the quadratic polynomial which satisfies the above conditions is x2 − 4x + 1.