# PolynomialsExercise – 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

Solution:

By the factorization method,

x2-4x+2x-8 = 0

x(x-4) +2(x-4) = 0

(x+2) (x-4) = 0

x = -2, 4

So, zeroes α=-2, β=4

Verification,

-2+4 = -(-2)/1

2 = 2

-2×4 = -8/1

-8 = -8

The relation between the coefficients is true in both the cases.

(ii) 4s2-4s+1

Solution:

By the factorization method,

4s2-2s-2s+1 = 0

2s(2s-1) -1(2s-1) = 0

(2s-1) (2s-1) = 0

s = 1/2, 1/2

So, zeroes α= 1/2, β= 1/2

Verification,

sum of zeroes(α+β) =(-b )/a

½ + 1/2 = -(-4/4)

2/2 = 4/4

1 = 1

product of zeroes (αβ) =(c )/a

½ × 1/2 = 1/4

1/4 = 1/4

The relation between the coefficients is true in both the cases.

(iii) 6x2-3-7x

Solution:

By the factorization method,

6x2-9x+2x-3 = 0

3x(2x-3) +1(2x-3) = 0

(3x+1) (2x-3) = 0

x = -1/3, 3/2

So, zeroes α= 3/2, β= -1/3

Verification,

sum of zeroes(α+β) =(-b )/a

3/2 + -1/3 = -(-7)/6

(9-2)/6 = 7/6

7/6 = 7/6

product of zeroes (αβ) =(c )/a

3/2 × -1/3 = -3/6

-3/6 = -3/6

The relation between the coefficients is true in both the cases.

(iv) 4u2+8u

Solution:

By the factorization method,

4u(u+2) = 0

u = 0, -2

So, zeroes α= 0, β= -2

Verification,

sum of zeroes (α+β) =(-b )/a

0 + -2 = -8/4

-2 = -2

product of zeroes (αβ) =(c )/a

0 × -2 = 0/4

0 = 0

The relation between the coefficients is true in both the cases.

(v) t2-15

Solution:

By the factorization method,

t2 - 15 = 0

t = ±√15

t = √15, -√15

So, zeroes α= √15, β= -√15

Verification,

sum of zeroes (α+β) =(-b )/a

√15 + -√15 = -(0)/1

0 = 0

product of zeroes (αβ) =(c )/a

√15 × -√15 = -15

-15 = -15

The relation between the coefficients is true in both the cases.

(vi) 3x2-x-4

Solution:

By the factorization method,

3x2-4x+3x-4 = 0

x(3x-4) +1(3x-4) = 0

(3x-4) (x+1) = 0

x = 4/3, -1

So, zeroes α= 4/3, β= -1

Verification,

sum of zeroes (α+β) =(-b )/a

4/3 + -1 = -(-1)/3

(4-3)/3 = 1/3

1/3 = 1/3

product of zeroes (αβ) =(c )/a

4/3 × -1 = -4/3

-4/3 = -4/3

The relation between the coefficients is true in both the cases.

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

(i) ¼, -1

Solution:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = ¼

-b/a = ¼

by comparison,

-b = 1, a = 4

b = -1, a = 4

product of zeros αβ = c/a = -1

c/a = -1

or                c/4 = -1  (a = 4)

c = -4

Now,

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

Hence,

ax2+bx+c

4x2-1x-4

(ii) √2, 1/3

Solution:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = √2

-b/a = √2

or            -b/a = √2/1

by comparison,

-b = √2, a = 1

b = -√2, a = 1

product of zeros αβ = c/a = 1/3

c/a = 1/3

or               c/1 = 1/3  (a = 1)

c = 1/3

Now,

b = -√2, a = 1, c = 1/3

Hence,

ax2+bx+c = 0

3x2-3√2+1 = 0

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = 0

-b/a = 0

or            -b/a = 0/1

by comparison,

b = 0, a = 1

product of zeros αβ = c/a = 5

c/a = 5/1

or               c/1 = √5/1  (a = 1)

c = 5

Now,

b = 0, a = 1,                                     c = √5

Hence,

ax2+bx+c = 0

x2-0x+√5 = 0

x2+√5 = 0

(iv) 1, 1

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = 1

-b/a = 1

or            -b/a = 1/1

by comparison,

b = -1, a = 1

product of zeros αβ = c/a = 1

c/a = 1/1

or               c/1 = 1/1  (a = 1)

c = 1

Now,

b = -1, a = 1,                                     c = 1

Hence,

ax2+bx+c = 0

x2-1x+1 = 0

(v) -1/4, 1/4

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = -1/4

-b/a = -1/4

or            b/a = 1/4

by comparison,

b = 1, a = 4

product of zeros αβ = c/a = 1/4

c/a = 1/4

or               c/4 = 1/4  (a = 4)

c = 1

Now,

b = 1, a = 4,                                     c = 1

Hence,

ax2+bx+c = 0

4x2+1x+1 = 0

(vi) 4, 1

हल:

Let a quadratic equation (polynomial) be ax2+bx+c and its zeroes are α, β.

According to Question,

sum of zeroes α + β = -b/a = 4

-b/a = 4

or            b/a = -4/1

by comparison,

b = -4, a = 1

product of zeros αβ = c/a = 1

c/a = 1

or               c/1 = 1/1  (a = 1)

c = 1

Now,

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

Hence,

ax2+bx+c = 0

x2-4x+1 = 0