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

Math is tough but solving them is impossible said no one ever!! Make use of these NCERT Solutions For Class 10 Maths for finding solutions and also note that NCERT Solutions for Class 10 Maths PDF are available online for extra reference. This chapter includes NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2. Practice more so you score more!!

## Exercise 2.2

Question 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 + 2x – 8
⇒ x(x – 4) + 2(x – 4)
⇒ (x – 4) (x + 2)
⇒ x – 4 = 0 or x + 2 = 0
⇒ x = 4 or x= -2
Verification: (ii) 4s2 – 4s + 1
⇒ 4s² – 2s – 2s + 1
⇒ 2s(2s – 1) – 1(2s – 1)
⇒ (2s – 1) (2s – 1)
2s – 1 = 0 or 2s – 1 = 0
⇒ s = $$\frac { 1 }{ 2 }$$ or s = $$\frac { 1 }{ 2 }$$
Verification: (iii) 6x2 – 3 – 7x
⇒ 6x² – 7x + 2x – 3
⇒ 3x(2x – 3) + 1(2x – 3)
⇒ (2x – 3) (3x + 1)
2x – 3 = 0 or 3x + 1 = 0
⇒ x = $$\frac { 3 }{ 2 }$$ or x = $$\frac { -1 }{ 3 }$$
Verification: (iv) 4u2 + 8u
⇒ 4u(u + 2)
⇒ u – 4 = 0 or u + 2 = 0
⇒ u = 4 or u= – 2
Verification: (v) t2 -15
⇒ t² = 15
⇒ t = ± $$\sqrt{15}$$
t = $$\sqrt{15}$$ or t = – $$\sqrt{15}$$
Verification: (vi) 3x2 – x – 4
⇒ 3x² – 4x + 3x – 4
⇒ x(3x – 4) + 1(3x – 4)
⇒ (3x – 4) (x + 1)
3x – 4 = 0 or x + 1 = 0
⇒ x = $$\frac { 4 }{ 3 }$$ or x = – 1
Verification: Question 2.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $$\frac { 1 }{ 4 }$$, – 1
(ii) $$\sqrt{2}$$, $$\frac { 1 }{ 3 }$$
(iii) 0, $$\sqrt{5}$$
(iv) 1, 1
(v) $$\frac { -1 }{ 4 }$$, $$\frac { 1 }{ 4 }$$
(vi) 4, 1
Solution:
(i) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = $$\frac { 1 }{ 4 }$$ = $$\frac { -b }{ a }$$
and α.β = – 1 = $$\frac { c }{ a }$$
If a = 4, then b = – 1 and c = – 4
So, one quadratic polynomial which fits the given condition is 4x² – x – 4.

(ii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = $$\sqrt{2}$$ = $$\frac { -b }{ a }$$
and α.β = $$\frac { 1 }{ 3 }$$ = $$\frac { c }{ a }$$
If a = 4, then b = – $$\sqrt{2}$$ and c = $$\frac { 1 }{ 3 }$$
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{2x}$$ + $$\frac { 1 }{ 3 }$$ or 3x² – 3$$\sqrt{2x}$$ + 1

(iii) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 0 = $$\frac { -b }{ a }$$
and α.β = $$\sqrt{5}$$ = $$\frac { c }{ a }$$
If a = 1, then b = 0 and c = $$\sqrt{5}$$
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{5}$$

(iv) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 1 = $$\frac { -b }{ a }$$
and α.β = 1 = $$\frac { c }{ a }$$
If a = 1, then b = – 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is x² – $$\sqrt{5}$$
1x² – 1x 1 = 0
or x² – x + 1 = 0

(v) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = –$$\frac { 1 }{ 4 }$$ = $$\frac { -b }{ a }$$
and α.β = $$\frac { 1 }{ 4 }$$ = $$\frac { c }{ a }$$
If a = 1, then b = 1 and c = 1
Now, put the values of a, b and c in equation ax² + bx + c
So, one quadratic polynomial which fits the given condition is 4x² + x + 1 = 0

(vi) Let the polynomial be ax² + bx + c and its zeroes be α and ß.
Then α + β = 4 = $$\frac { -b }{ a }$$
and α.β = 1 = $$\frac { c }{ a }$$
If a = 1, then b = – 4 and c = 1
So, one quadratic polynomial which fits the given condition is x² – 4x + 1 = 0

We hope you found these NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 helped you with finding the solutions easily.

Make sure to practice the remaining parts of the Class 10 Maths NCERT Solutions English Medium Chapter 2.

error: Content is protected !!