NCERT Solutions Maths Class 10 Chapter 2

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## NCERT Solutions Maths Class 10 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

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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = 1/4 = −b/a

And product of zeroes = α × β = −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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = √2 = 3√2/3 = −b/a

And product of zeroes = α × β = 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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = 0 = 0/1 = −b/a

And product of zeroes = α × β = √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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = 1 = −(−1)/1 = −b/a

And product of zeroes = α × β = 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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = −1/4 = −b/a

And product of zeroes = α × β = 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 α and β be two zeroes of above quadratic polynomial.

Then, sum of zeroes = α + β = 4  = −(−4)/1 = −b/a

And product of zeroes = α × β = 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.