# Class10 NCRT Polynomials Exercise – 2.2 pdf || UP Board

__Polynomials__

__Exercise – 2.2__

__Polynomials__

__Exercise – 2.2__

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

**(****i****) ****x ^{2}**

**-2**

**x**

**-8**

**Solution****:**

By the
factorization method,

x^{2}-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****) ****4s ^{2}-4s+1**

**Solution****:**

By the
factorization method,

4s^{2}-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****) ****6x ^{2}**

**-**

**3**

**-**

**7x**

**Solution****:**

By the
factorization method,

6x^{2}-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****) ****4u ^{2}+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****) ****t ^{2}-15**

**Solution:**

By the
factorization method,

t^{2} - 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****) ****3x ^{2}**

**-**

**x**

**-**

**4**

**Solution:**

By the
factorization method,

3x^{2}-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 ax^{2}+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,

ax^{2}+bx+c

4x^{2}-1x-4

The
quadratic polynomial is 4x^{2}-1x-4.

**(****ii****) ****√2****,
1/3**

**Solution****:**

Let a
quadratic equation (polynomial) be ax^{2}+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,

ax^{2}+bx+c
= 0

3x^{2}-3√2+1
= 0

The quadratic polynomial is 3x^{2}-3√2+1.

**हल****:**

Let a
quadratic equation (polynomial) be ax^{2}+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,

ax^{2}+bx+c
= 0

x^{2}-0x+√5 = 0

x^{2}+√5
= 0

The
quadratic polynomial is x^{2}+√5.

**(****iv****) ****1****,
****1**

**हल****:**

Let a
quadratic equation (polynomial) be ax^{2}+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,

ax^{2}+bx+c
= 0

x^{2}-1x+1 = 0

The
quadratic polynomial is x^{2}-1x+1.

**(****v****) ****-1/4****,
****1/4**

**हल****:**

Let a
quadratic equation (polynomial) be ax^{2}+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,

ax^{2}+bx+c
= 0

4x^{2}+1x+1 = 0

The
quadratic polynomial is 4x^{2}+1x+1.

**(****vi****) ****4****,
****1**

**हल****:**

Let a
quadratic equation (polynomial) be ax^{2}+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,

ax^{2}+bx+c
= 0

x^{2}-4x+1 = 0

The
quadratic polynomial is x^{2}-4x+1.

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