These NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.
NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
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