Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 2 Polynomials Ex 2.2 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 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) Given Quadratic polynomial,
x2 – 2x – 8
∵ [S = -2, P = -8]
= x2 – 4x + 2x – 8
= x (x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero,
iff (x – 4) = 0 or (x + 2) = 0
iff x = 4 or x = – 2
Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.
Now, Sum of zeroes = (- 2) + (4) = 2
= \(\frac{-(-2)}{1}=-\frac{(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 2) (4) = – 8
= \(\frac{-8}{1}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficient are verified.
(ii) Given quadratic polynomial,
4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
∵ [S = -4, P = 4 × 1]
= 2s (2s – 1) – 1 (2s – 1)
= (2s – 1) (2s – 1)
The value of 4s2 – 4s + 1 is zero
iff (2s – 1) = 0 or (2s – 1) = 0
iff s = \(\frac{1}{2}\) or s = \(\frac{1}{2}\)
Therefore, zeroes of 4s2 – 4s + 1 are \(\frac{1}{2}\) and \(\frac{1}{2}\)
Now, sum of zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1
= \(\frac{-(-4)}{4}=\frac{-(\text { Coefficient of } s)}{\left(\text { Coefficient of } s^{2}\right)}\)
Product of Zeroes = \(\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}\)
= \(\frac{\text { Constant term }}{\text { Coefficient of } s^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.
(iii) Given quadratic polynomial,
6x2 -3 – 7x
= 6x2 – 7x – 3
∵ [S = – 7, P = 6x-3=-18]
= 6x2 -9x + 2x-3
= 3x (2x – 3) + 1 (2x – 3)
= (2x – 3) (3x + 1)
The value of 6x2 – 3 – 7x is zero
iff (2x – 3) = 0 or 3x + 1 = 0
iff x = \(\frac{3}{2}\) or x = –\(\frac{1}{3}\)
Therefore, zeroes of 6x2 – 3 – 7x are \(\frac{3}{2}\) and –\(\frac{1}{3}\)
Now, Sum of zeroes = \(\frac{1}{3}\)
= \(\frac{3}{2}+\left(\frac{-1}{3}\right)\)
= \(\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}\)
= \(\frac{7}{6}=\frac{-(-7)}{6}\)
= \(\frac{-(\text { Coefficient of } x)}{\text { Coefficient of } x^{2}}\)
Product of zeroes = \(\left(\frac{3}{2}\right)\left(\frac{-1}{3}\right)\)
= \(\frac{-3}{6}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.
(iv) Given quadratic polynomial,
4u2 + 8u = 4u(u + 2)
The value of 4u2 + 8 u is zero
iff 4u = 0 or u + 2 = 0
iff u = 0 or u = – 2
Therefore, zeroes of 4u2 + 8M are 0 and – 2
Now, Sum of zeroes = 0 + (- 2) = -2
= \(\frac{-8}{4}\)
= \(-\frac{(\text { Coefficient of } u)}{\text { Coefficient of } u^{2}}\)
Product of zeroes = (0) (- 2) = 0
= \(\frac{0}{4}=\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.
(v) Given quadratic polynomial,
t2 – 15 = t2 – (√15)2
= (t – √15) (t + √15)
The value of t2 – 15 is zero
iff t – √15 = 0 or t + √15 = 0
iff t = √15 or t = – √15
Therefore, zeroes of t2 – 15 are – √15 and √15.
Now, Sum of zeroes = -√15 + √15 = 0 = \(\frac{0}{1}\)
= \(\frac{-(\text { Coefficient of } t)}{\text { Coefficient of } t^{2}}\)
Product of zeroes = (-√15) (√15) = – 15 = \(-\frac{15}{1}\)
= \(\frac{0}{1}\)
= \(=\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.
(vi) Given quadratic polynomial,
3x2 – x – 4
= 3x2 + 3x – 4x – 4
= 3x (x + 1) – 4 (x + 1)
∵ [S = – 1, P = 3 x – 4 = – 12]
= (x + 1) (3x – 4)
The value of 3x2 – x – 4 is zero
iff (x + 1) = 0 or 3x – 4 = 0
iff x = -1 or x = \(\frac{4}{3}\)
Therefore, zeroes of 3x2 – x – 4 are – 1 and \(\frac{4}{3}\)
Now, Sum of zeroes = – 1 + \(\frac{4}{3}\)
= \(\frac{-3+4}{3}=\frac{1}{3}\)
= \(\frac{-(-1)}{3}=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\)
Product of zeroes = (- 1) \(\frac{4}{3}\)
= \(\frac{-4}{3}=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\)
Hence, relationship between the zeroes and the coefficients are verified.
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) √2, \(\frac{1}{3}\)
(iii) 0, √5
(iv) 1, 1
(v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
(vi) 4, 1. [MQP 2015]
Solution:
(i) Given that, sum of zeroes and products of zeroes of given polynomial \(\frac{1}{4}\) are -1 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = \(\frac{1}{4}\)
and αβ = Product of zeroes = – 1
Now, ax2 + bx + c = k (x – α) (x – β) where k is any constant
= k [x2 – (α + β)x + αβ]
= k[x2 – \(\frac{1}{4}\)x + (-1)]
= k[x2 – \(\frac{1}{4}\)x – 1]
for different value of k, we get different quadratic polynomials.
(ii) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are √2 and \(\frac{1}{3}\) respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = √2
and αβ = Product of zeroes = \(\frac{1}{3}\)
Now, ax2 + bx + c = k (x – α) (x – β) where k is any constant
= k[x2 – (α + β) x + αβ]
= k[x2 – √2x + \(\frac{1}{3}\)]of k, we get different quadratic polynomial.
for different values of k, we get different quadratic polynomial.
(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 0
and αβ = Product of zeroes = √5
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant
= k [x2 – (a + (α+ β)x + αβ)
= k[x2 – 0x + √5]
= k[x2 + √5]
for different values of k, we get different quadratic polynomial.
(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 1 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant.
= k [x2 – (α + β)x + αβ]
= k [x2 – 0x + √5]
= it [x2 – x + √5]
for different values of k, we get different quadratic polynomial.
(v) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are \(-\frac{1}{4}\) and \(\frac{1}{4}\) respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = –\(\frac{1}{4}\)
and αβ = Product of zeroes = \(\frac{1}{4}\)
Now ax2 + bx + c = k (x – α) (x – β) where k is any constant
= k[x2 – (α + β) x + αβ]
= k[x2 – (-\(\frac{1}{4}\))x + \(\frac{1}{4}\)]
= k[x2 + \(\frac{1}{4}\)x + \(\frac{1}{4}\)]
For different values of k, we get different quadratic polynomial.
(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be
ax2 + bx + c and its zeroes be α and β
α + β = sum of zeroes = 4 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant
= k [x2 – (α + β) x + αβ]
= k [x2 – 4x + 1]
For different values of k, we get different quadratic polynomials.