# RD Sharma Solutions Class 10 Polynomials Exercise 2.1

## RD Sharma Solutions Class 10 Chapter 2 Exercise 2.1

### RD Sharma Class 10 Solutions Chapter 2 Ex 2.1 PDF Free Download

#### Exercise 2.1

Q.1: Find the zeroes of each of the following quadratic polynomials and verify the relationship between the zeroes and their coefficients:

(i) $f\left ( x \right ) = x^{2} – 2x – 8$

(ii) $g\left ( s \right ) = 4s^{2} – 4s + 1$

(iii) $6x^{2} – 3 – 7x$

(iv) $h\left ( t \right ) = t^{2} – 15$

(v) $p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$

(vi) $q\left ( x \right ) = \sqrt{3}x^{2} + 10x + 7\sqrt{3}$

(vii) $f\left ( x \right ) = x^{2} – \left ( \sqrt{3} + 1 \right )x + \sqrt{3}$

(viii) $g\left ( x \right ) = a\left ( x^{2} + 1 \right ) – x\left ( a^{2} + 1 \right )$

Solution:

(i) $f\left ( x \right ) = x^{2} – 2x – 8$

We have,

$f\left ( x \right ) = x^{2} – 2x – 8$

= $x^{2} – 4x + 2x – 8$

= $x\left ( x – 4 \right ) + 2\left ( x – 4 \right )$

= $\left ( x + 2 \right )\left ( x – 4 \right )$

Zeroes of the polynomials are -2 and 4.

Now,

Sum of the zeroes = $\frac{- \; coefficient \; of \; x}{coefficient \; of \; x}$

-2 + 4 = $\frac{-\left ( -2 \right )}{1}$

2 = 2

Product of the zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$

-8 = $\frac{-8}{1}$

-8 = -8

Hence, the relationship is verified.

(ii) $g\left ( s \right ) = 4s^{2} – 4s + 1$

We have,

$g\left ( s \right ) = 4s^{2} – 4s + 1$

= $4s^{2} – 2s – 2s + 1$

= $2s\left ( 2s – 1 \right ) -1\left ( 2s – 1 \right )$

= $\left ( 2s – 1 \right )\left ( 2s – 1 \right )$

Zeroes of the polynomials are $\frac{1}{2}$ and $\frac{1}{2}$.

Sum of zeroes = $\frac{- coefficient \; of \; s}{coefficient \; of \; s^{2}}$

$\frac{1}{2} + \frac{1}{2} = \frac{-\left ( -4 \right )}{4}$

1 = 1

Product of zeroes =  $\frac{constant \; term}{Coefficient \; of \; x}$

$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

$\frac{1}{4} = \frac{1}{4}$

Hence, the relationship is verified.

(iii) $6s^{2}-3-7x$

= $6s^{2}-7x-3$ = (3x + 11)(2x – 3)

Zeros of the polynomials are $\frac{3}{2} \; and \; \frac{-1}{3}$

Sum of the zeros = $\frac{- coefficient \; of \; x}{coefficient of x^{2}}$

$\frac{-1}{3} + \frac{3}{2} = \frac{-\left ( -7 \right )}{6}$

$\frac{7}{6} = \frac{7}{6}$

Product of the zeroes = $\frac{constant \; term}{coefficient \; of \; x^{2}}$

$\frac{-1}{3} \times \frac{3}{2} = \frac{-3}{6}$

$\frac{-3}{6} = \frac{-3}{6}$

Hence, the relationship is verified.

(iv) $h\left ( t \right ) = t^{2} – 15$

We have,

$h\left ( t \right ) = t^{2} – 15$

= $t^{2} – \sqrt{15}$

= $\left ( t + \sqrt{15} \right )\left ( t – \sqrt{15} \right )$

Zeroes of the polynomials are $-\sqrt{15} \; and \; \sqrt{15}$

Sum of the zeroes = 0

$-\sqrt{15} + \sqrt{15}$ = 0

0 = 0

Product of zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$ = $\frac{-15}{1}$

$-\sqrt{15} \times \sqrt{15} = -15$

-15 = -15

Hence, the relationship verified.

(v) $p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$

We have,

$p\left ( x \right ) = x^{2} + 2\sqrt{2}x – 6$

= $x^{2} + 3\sqrt{2}x + 3\sqrt{2}x – 6$

= $x\left ( x + 3\sqrt{2} \right ) – \sqrt{2}\left ( x + 3\sqrt{2} \right )$

= $\left ( x + 3\sqrt{2} \right )\left ( x – \sqrt{2} \right )$

Zeroes of the polynomials are $3\sqrt{2}$ and $- 3\sqrt{2}$

Sum of the zeroes = $\frac{-2\sqrt{2}}{1}$

$\sqrt{2} – 3\sqrt{2} = -2\sqrt{2}$

$– 2\sqrt{2} = -2\sqrt{2}$

Product of the zeroes = $\frac{constant \; term}{Coefficient \; of \; x}$

$\sqrt{2} \times -3\sqrt{2} = \frac{-6}{1}$

-6 = -6

Hence, the relationship is verified.

(vi) q(x) = $\sqrt{3}x^{2}+10x+7\sqrt{3}$

= $\sqrt{3}x^{2}+7x+3x+7\sqrt{3}$

= $\sqrt{3}x\left( x +\sqrt{3}\right) 7\left( x + sqrt{3} \right)$

= $\left( x +\sqrt{3}\right) \left( 7 + sqrt{3} \right)$

Zeros of the polynomials are $-\sqrt{3}and \frac {-7}{\sqrt{3}}$

Sum of zeros = $\frac {-10}{\sqrt{3}}$

$-\sqrt{3}- \frac {7}{\sqrt{3}} = \frac {-10}{\sqrt{3}}$

$\frac {-10}{\sqrt{3}} = \frac {-10}{\sqrt{3}}$

Product of the polynomials are $-\sqrt{3}, \frac {-7}{\sqrt{3}}$

7 = 7

Hence, the relationship is verified.

(vii)  h(x) = $x^{2}-\left( \sqrt{3}+1\right)x + \sqrt{3}$

= $x^{2}-\sqrt{3} -x + \sqrt{3}$

=$x(x – \sqrt{3}) -1(x-\sqrt{3})$

= $(x – \sqrt{3}) (x-1)$

Zeros of the polynomials are 1 and $\sqrt{3}$

Sum of zeros = $\frac{-co\;efficient \;of \;x}{co\; efficient \;of\; x^{2}} = -[-\sqrt{3}-1]$

$1+\sqrt{3} = \sqrt{3} +1$

Product of zeros = $\frac{-co \; efficient \; of \; x}{co \; efficient \; of \; x^{2}} = \sqrt{3}$

$\sqrt{3} = \sqrt{3}$

Hence, the relationship is verified

(viii) g(x) = $a\left [ \left ( x^{2} + 1 \right ) – x\left ( a^{2} + 1 \right ) \right ]^{2}$

= $ax^{2}+a-a^{2}x -x$

= $ax^{2}-[(a^{2}x +1)]+a$

= $ax^{2}-a^{2}x –x +a$

= $ax(x-a)-1(x –a)$ = $\left ( x – a \right )\left ( ax – 1 \right )$

Zeros of the polynomials are $\frac{1}{a} \; and \; 1$

Sum of the zeros = $\frac{a[-a^{2}-1]}{a}$

$\frac{1}{a}+a=\frac{a^{2}+1}{a}\ \frac{a^{2}+1}{a}=\frac{a^{2}+1}{a}$

Product of zeros = a/a

$\frac{1}{a}\times a= \frac{a}{a}$

1 = 1

Hence, the relationship is verified.

Q.2: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – 5x + 4$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$.

Solution: We have,

$\alpha \; and \; \beta$ are the roots of the quadratic polynomial.

$f\left ( x \right ) = x^{2} – 5x + 4$

Sum of the roots = $\alpha \; +\; \beta$ = 5

Product of the roots = $\alpha \beta$ = 4

So,

$\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$ = $\frac{\beta + \alpha }{\alpha \beta } – 2\alpha \beta$

= $\frac{5}{4} – 2\times 4 = \frac{5}{4} – 8 = \frac{-27}{4}$

Q.3: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – 5x + 4$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta } – 2 \alpha \beta$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.

$p\left ( y \right ) = x^{2} – 5x + 4$

Sum of the zeroes = $\alpha \; +\; \beta$ = 5

Product of the roots = $\alpha \beta$ = 4

So,

$\frac{1}{\alpha } + \frac{1}{\beta } – 2 \alpha \beta$

= $\frac{ \beta + \alpha – 2 \alpha ^{2} \beta ^{2}}{\alpha \beta }$

= $\frac{ \left ( \alpha + \beta \right ) – 2 \left ( \alpha \beta \right )^{2}}{\alpha \beta }$

= $\frac{ \left ( 5 \right ) – 2 \left ( 4 \right )^{2}}{4}$

= $\frac{ 5 – 2 \times 16}{4}$ = $\frac{5 – 32}{4}$ = $\frac{- 27}{4}$

Q.4: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( y \right ) = 5y^{2} – 7y + 1$, find the value of $\frac{1}{\alpha } + \frac{1}{\beta }$.

Solution: Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.

$p\left ( y \right ) = 5y^{2} – 7y + 1$

Sum of the zeroes = $\alpha \; +\; \beta$ = 7

Product of the roots = $\alpha \beta$ = 1

So,

$\frac{1}{\alpha } + \frac{1}{\beta }$ = $\frac{\alpha + \beta}{\alpha \beta}$ = $\frac{7}{1}$ = 7

Q.5: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – x – 4$, find the value of $\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.

We have,

$f\left ( x \right ) = x^{2} – x – 4$

Sum of zeroes = $\alpha \; +\; \beta$ = 1

Product of the zeroes = $\alpha \beta$ = -4

So,

$\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$ = $\frac{\alpha + \beta }{\alpha \beta } – \alpha \beta$

= $\frac{1}{-4} – \left ( -4 \right ) \; = \frac{-1}{4} + 4$

= $\frac{-1 + 16}{4}$ = $\frac{15}{4}$

Q.6: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} + x – 2$, find the value of $\frac{1 }{\alpha } – \frac{1 }{\beta }$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial.

We have,

$f\left ( x \right ) = x^{2} + x – 2$

Sum of zeroes = $\alpha \; +\; \beta$ = 1

Product of the zeroes = $\alpha \beta$ = -2

So,

$\frac{1 }{\alpha } + \frac{1 }{\beta } – \alpha \beta$ = $\frac{\alpha + \beta }{\alpha \beta }$

= $\frac{\beta – \alpha }{\alpha \beta }$

$= \frac{\beta – \alpha }{\alpha \beta }\times \frac{\left ( \alpha – \beta \right )}{\alpha \beta }$

$= \frac{\sqrt{\left ( \alpha + \beta \right )^{2} – 4\alpha \beta }}{\alpha \beta }$

$= \frac{\sqrt{1 + 8}}{2} = \frac{\sqrt{9}}{2} = \frac{3}{2}$

Q.7: If one of the zero of the quadratic polynomial $f\left ( x \right ) = 4x^{2} – 8kx – 9$ is negative of the other, then find the value of k.

Solution:

Let, the two zeroes of the polynomial $f\left ( x \right ) = 4x^{2} – 8kx – 9$ be $\alpha \; and \; -\alpha$.

Product of the zeroes = $\alpha \times -\alpha$ = -9

Sum of the zeroes = $\alpha + \left (-\alpha \right )$ = -8k = 0                              $Since, \alpha – \alpha = 0$

$\Rightarrow 8k = 0$

$\Rightarrow k = 0$

Q.8: If the sum of the zeroes of the quadratic polynomial $f\left ( t \right ) = kt^{2}+ 2t + 3k$ is equal to their product, then find the value of k.

Solution: Let the two zeroes of the polynomial $f\left ( t \right ) = kt^{2}+ 2t + 3k$ be $\alpha \; and \; \beta$.

Sum of the zeroes = $\alpha + \beta$ = 2

Product of the zeroes = $\alpha \times \beta$ = 3k

Now,

$\frac{-2}{k} = \frac{3k}{k}$

$\Rightarrow 3k = -2$

$\Rightarrow k = \frac{-2}{3}$

So, k = 0 and $\Rightarrow k = \frac{-2}{3}$

Q.9: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( x \right ) = 4x^{2} – 5x – 1$, find the value of $\alpha ^{2} \beta + \alpha \beta ^{2}$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $p\left ( x \right ) = 4x^{2} – 5x – 1$

So, Sum of the zeroes = $\alpha + \beta$ = $\frac{5}{4}$

Product of the zeroes = $\alpha \times \beta$$\frac{-1}{4}$

Now,

$\alpha ^{2} \beta + \alpha \beta ^{2}$ = $\alpha \beta \left ( \alpha + \beta \right )$

= $\frac{5}{4} \left ( \frac{-1}{4} \right )$

= $\frac{-5}{16}$

Q.10: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( t \right ) = t^{2} – 4t + 3$, find the value of $\alpha ^{4} \beta ^{3} + \alpha ^{3} \beta ^{4}$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( t \right ) = t^{2} – 4t + 3$

So, Sum of the zeroes = $\alpha + \beta$ = 4

Product of the zeroes = $\alpha \times \beta$ =  3

Now,

$\alpha ^{4} \beta ^{3} + \alpha ^{3} \beta ^{4}$ = $\alpha ^{3} \beta ^{3} \left ( \alpha + \beta \right )$

= $\left ( 3 \right )^{3} \left ( 4 \right )$ = 108

Q.11: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$, find the value of $\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$.

Sum of the zeroes =  $\alpha + \beta$ = $\frac{-1}{6}$

Product of the zeroes = $\alpha \times \beta$$\frac{-1}{3}$

Now,

$\frac{\alpha }{\beta } + \frac{\beta }{\alpha }$

= $\frac{\left (\alpha ^{2} + \beta ^{2} \right ) – 2\alpha \beta}{\alpha \beta }$

By substitution the values of the sum of zeroes and products of the zeroes, we will get

= $\frac{-25}{12}$

Q.12: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$, find the value of $\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = 6x^{2} + x – 2$.

Sum of the zeroes = $\alpha + \beta$ = $\frac{6}{3}$

Product of the zeroes = $\alpha \times \beta$$\frac{4}{3}$

Now,

$\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$.

= $\frac{\alpha ^{2} + \beta ^{2}}{\alpha \beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$

= $\frac{\left ( \alpha + \beta \right )^{2} – 2\alpha \beta }{\alpha \beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$

By substituting the values of sum and product of the zeroes, we will get

$\frac{\alpha }{\beta } + 2\left [ \frac{1}{\alpha } + \frac{1}{\beta } \right ] + 3\alpha \beta$ = 8

Q.13: If the squared difference of the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} + px + 45$ is equal to 144, find the value of p.

Solution:

Let the two zeroes of the polynomial be $\alpha \; and \; \beta$.

We have,

$f\left ( x \right ) = x^{2} + px + 45$

Now,

Sum of the zeroes = $\alpha + \beta$ =-p

Product of the zeroes = $\alpha \times \beta$ =  45

So,

$\left ( \alpha + \beta \right )^{2} – 4\alpha \beta = 144$

$\left (p\right )^{2} – 4\times 45 = 144$

$\left (p\right )^{2} = 144 + 180$

$\left (p\right )^{2} = 324$

$p = \sqrt{324}$

$p = \pm 18$

Thus, in the given equation, p will be either 18 or -18.

Q.14: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – px + q$,  prove that $\frac{\alpha ^{2}}{\beta ^{2}} + \frac{\beta ^{2}}{\alpha ^{2}} = \frac{p^{4}}{q^{2}} – \frac{4p^{2}}{q} + 2$.

Solution:

Since, $\alpha \; and \; \beta$ are the roots of the quadratic polynomial given in the question.

$f\left ( x \right ) = x^{2} – px + q$

Now,

Sum of the zeroes = p = $\alpha + \beta$

Product of the zeroes = q = $\alpha \times \beta$

LHS = $\frac{\alpha ^{2}}{\beta ^{2}} + \frac{\beta ^{2}}{\alpha ^{2}}$

= $\frac{\alpha ^{4} + \beta ^{4}}{\alpha ^{2} \beta ^{2}}$

= $\frac{\left (\alpha ^{2} + \beta ^{2} \right )^{2} – 2\left (\alpha \beta \right )^{2}}{\left (\alpha \beta \right )^{2}}$

= $\frac{\left [\left (\alpha + \beta \right )^{2} – 2\alpha \beta \right ]^{2} – 2\left (\alpha \beta \right )^{2}}{\left (\alpha \beta \right )^{2}}$

= $\frac{\left [\left (p \right )^{2} – 2 q \right ]^{2} – 2\left (q \right )^{2}}{\left (q \right )^{2}}$

= $\frac{\left (p ^{4} + 4 q ^{2} – 4 p ^{2} q \right ) – 2q^{2}}{q ^{2}}$

= $\frac{p ^{4} + 2 q ^{2} – 4 p ^{2} q }{q ^{2}}$

= $\frac{p ^{2}}{q ^{2}} + 2 – \frac{4 p^{2}}{q}$

= $\frac{p ^{2}}{q ^{2}} – \frac{4 p^{2}}{q} + 2$

LHS = RHS

Hence, proved.

Q.15: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x^{2} – p( x + 1 ) – c$, show that $\left ( \alpha + 1 \right )\left ( \beta + 1 \right ) = 1 – c$.

Solution:

Since, $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial

$f\left ( x \right ) = x^{2} – p( x + 1 ) – c$

Now,

Sum of the zeroes = $\alpha + \beta$ = p

Product of the zeroes = $\alpha \times \beta$ = (- p – c)

So,

$\left (\alpha + 1 \right )\left ( \beta + 1 \right )$

= $\alpha \beta + \alpha + \beta + 1$

= $\alpha \beta + \left (\alpha + \beta \right ) + 1$

= $\left (- p – c \right ) + p + 1$

= 1 – c = RHS

So, LHS = RHS

Hence, proved.

Q.16: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial such that $\alpha + \beta = 24 \; and \; \alpha – \beta = 8$, find a quadratic polynomial having $\alpha \; and \; \beta$ as its zeroes.

Solution:

We have,

$\alpha + \beta = 24$ …………E-1

$\alpha – \beta = 8$ ………….E-2

By solving the above two equations accordingly, we will get

$2\alpha = 32$

$\alpha = 16$

Substitute the value of $\alpha$, in any of the equation. Let we substitute it in E-2, we will get

$\beta = 16 – 8$

$\beta = 8$

Now,

Sum of the zeroes of the new polynomial = $\alpha + \beta$ = 16 + 8 = 24

Product of the zeroes = $\alpha \beta$ = $16 \times 8$ = 128

K $x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$ = $x^{2} – 24 x + 128$

Hence, the required quadratic polynomial is $f\left ( x \right ) = x^{2} + 24 x + 128$

Q.17: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} – 1$, find a quadratic polynomial whose zeroes are $\frac{2\alpha }{\beta } \; and \; \frac{2\beta }{\alpha }$.

Solution:

We have,

$f\left ( x \right ) = x ^{2} – 1$

Sum of the zeroes = $\alpha + \beta$ = 0

Product of the zeroes = $\alpha \beta$ = -1

From the question,

Sum of the zeroes of the new polynomial = $\frac{2\alpha }{\beta } \; and \; \frac{2\beta }{\alpha }$

= $\frac{2\alpha ^{2} + 2\beta ^{2}}{\alpha \beta }$

= $\frac{2 \left ( \alpha ^{2} + \beta ^{2} \right )}{ \alpha \beta }$

= $\frac{2\left ( \left ( \alpha + \beta \right )^{2} – 2\alpha \beta \right )}{\alpha \beta }$

= $\frac{2\left ( 2 \right )1}{-1}$ { By substituting the value of the sum and products of the zeroes }

As given in the question,

Product of the zeroes = $\frac{\left ( 2 \alpha \right ) \left ( 2 \beta \right )}{\alpha \beta }$ = $\frac{4 \alpha \beta }{\alpha \beta }$ = 4

$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$

= k$x^{2} – \left ( -4 \right )x + 4$ = $x^{2} + 4x + 4$

Hence, the required quadratic polynomial is $f\left ( x \right ) = x^{2} + 4x + 4$

Q.18: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} – 3x – 2$, find a quadratic polynomial whose zeroes are $\frac{1}{2\alpha + \beta } \; and \; \frac{1}{2\beta + \alpha }$.

Solution:

We have,

$f\left ( x \right ) = x ^{2} – 3x – 2$

Sum of the zeroes = $\alpha + \beta$ = 3

Product of the zeroes = $\alpha \beta$ = -2

From the question,

Sum of the zeroes of the new polynomial  = $\frac{1}{2\alpha + \beta } + \frac{1}{2\beta + \alpha }$

= $\frac{2\beta + \alpha + 2\alpha + \beta }{\left ( 2\alpha + \beta \right ) \left ( 2\beta + \alpha \right )}$

= $\frac{3 \alpha + 3 \beta }{2 \left ( \alpha ^{2} + \beta ^{2} \right ) + 5\alpha \beta }$

= $\frac{3 \times 3}{ 2\left [ 2\left ( \alpha + \beta \right )^{2} – 2\alpha \beta + 5 \times \left ( -2 \right ) \right ]}$

= $\frac{9}{2\left [ 9 – \left ( -4 \right ) \right ] – 10}$

= $\frac{9}{2\left [ 13 \right ] – 10}$

= $\frac{9}{26 – 10}$ = $\frac{9}{16}$

Product of the zeroes = $\frac{1}{2 \alpha + \beta } \times \frac{1}{2\beta + \alpha }$

= $\frac{1}{\left ( 2 \alpha + \beta \right )\left ( 2 \beta + \alpha \right )}$

= $\frac{1}{4 \alpha \beta + 2 \alpha ^{2} + 2\beta ^{2} + \alpha \beta }$

= $\frac{1}{5 \alpha \beta + 2 \left (\alpha ^{2} + \beta ^{2} \right ) }$

= $\frac{1}{5 \alpha \beta + 2 \left (\left (\alpha + \beta \right )^{2} – 2\alpha \beta \right ) }$

= $\frac{1}{5 \times \left ( -2 \right ) + 2 \left (\left (3 \right )^{2} – 2\times \left ( -2 \right ) \right ) }$

= $\frac{1}{-10 + 26}$ = $\frac{1}{16}$

$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$

= k $\left ( x^{2} + \frac{9}{16}x + \frac{1}{16} \right )$

Hence, the required quadratic polynomial is k $\left ( x^{2} + \frac{9}{16}x + \frac{1}{16} \right )$.

Q.19: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial $f\left ( x \right ) = x ^{2} +px + q$, form a polynomial whose zeroes are $\left ( \alpha + \beta \right )^{2} \; and \; \left ( \alpha – \beta \right )^{2}$.

Solution:

We have,

$f\left ( x \right ) = x ^{2} + px + q$

Sum of the zeroes = $\alpha + \beta$ = -p

Product of the zeroes = $\alpha \beta$ = q

From the question,

Sum of the zeroes of new polynomial = $\left ( \alpha + \beta \right )^{2} + \left ( \alpha – \beta \right )^{2}$

= $\left ( \alpha + \beta \right )^{2} + \alpha ^{2} + \beta ^{2} – 2 \alpha \beta$

= $\left ( \alpha + \beta \right )^{2} + \left (\alpha + \beta \right )^{2} – 2\alpha \beta – 2 \alpha \beta$

= $\left ( -p \right )^{2} + \left (-p \right )^{2} – 2\times q – 2 \times q$

= $p^{2} + p^{2} – 4q$

= $2p^{2} – 4q$

Product of the zeroes of new polynomial = $\left ( \alpha + \beta \right )^{2}\left ( \alpha – \beta \right )^{2}$

= $\left ( -p \right )^{2} \left ( \left (-p \right ) ^{2} – 4q \right )$

= $p^{2} \left ( p^{2} – 4q \right )$

So, the quadratic polynomial is ,

$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$

= $x^{2} – \left ( 2p ^{2} – 4q \right )x + p^{2} \left ( p^{2} – 4q \right )$

Hence, the required quadratic polynomial is $f\left ( x \right ) = k \left (x^{2} – \left ( 2p ^{2} – 4q \right )x + p^{2} \left ( p^{2} – 4q \right ) \right )$.

Q.20: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial   $f\left ( x \right ) = x^{2} – 2x + 3$, find a polynomial whose roots are:

(i) $\alpha + 2 , \beta + 2$

(ii) $\frac{\alpha – 1}{\alpha + 1} , \frac{\beta – 1}{\beta + 1}$.

Solution:

We have,

$f\left ( x \right ) = x ^{2} – 2x + 3$

Sum of the zeroes = $\alpha + \beta$ = 2

Product of the zeroes = $\alpha \beta$ = 3

(i) Sum of the zeroes of new polynomial =  $\left ( \alpha + 2 \right ) + \left ( \beta + 2 \right )$

=   $\alpha + \beta + 4$

=   2 + 4 = 6

Product of the zeroes of new polynomial = $\left ( \alpha + 1 \right ) \left ( \beta + 1 \right )$

= $\alpha \beta + 2\alpha + 2\beta + 4$

= $\alpha \beta + 2\left (\alpha + \beta \right ) + 4$ = $3 + 2\left (2 \right ) + 4$ = 11

$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$

= $x^{2} – 6x + 11$

Hence, the required quadratic polynomial is $f\left ( x \right ) = k\left (x^{2} – 6x + 11 \right )$

(ii) Sum of the zeroes of new polynomial = $\frac{\alpha – 1}{\alpha + 1} + \frac{\beta – 1}{\beta + 1}$

= $\frac{\left ( \alpha – 1 \right )\left ( \beta + 1 \right ) + \left ( \beta – 1 \right )\left ( \alpha + 1 \right )}{\left ( \alpha + 1 \right )\left ( \beta + 1 \right )}$

= $\frac{\alpha \beta + \alpha – \beta – 1 + \alpha \beta + \beta – \alpha – 1}{\left ( \alpha + 1 \right )\left ( \beta + 1 \right )}$

= $\frac{3 – 1 + 3 – 1}{ 3 + 1 + 2}$ = $\frac{4}{6} = \frac{2}{3}$

Product of the zeroes of new polynomial = $\frac{\alpha – 1}{\alpha + 1} + \frac{\beta – 1}{\beta + 1}$

= $\frac{2}{6} = \frac{1}{3}$

$x^{2} – \left ( sum \; of \; the \; zeroes \right )x + \left ( product \; of \; the \; zeroes \right )$

= $x^{2} – \frac{2}{3}x + \frac{1}{3}$

Thus, the required quadratic polynomial is $f\left ( x \right ) = k\left (x^{2} – \frac{2}{3}x + \frac{1}{3} \right )$.

Q.21: If $\alpha \; and \; \beta$ are the zeroes of the quadratic polynomial  $f\left ( x \right ) = ax^{2} + bx + c$, then evaluate:

(i) $\alpha – \beta$

(ii) $\frac{1}{\alpha } – \frac{1}{\beta }$

(ii) $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$

(iv) $\alpha ^{2} \beta + \alpha \beta ^{2}$

(v) $\alpha ^{4} + \beta ^{4}$

(vi) $\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}$

(vii) $\frac{\beta }{a\alpha + b} + \frac{\alpha }{a\beta + b}$

(viii) $a\left [ \frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha } \right ] + b\left [ \frac{\alpha }{a} + \frac{\beta }{a} \right ]$

Solution:

$f\left ( x \right ) = ax^{2} + bx + c$

Here,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

Since, $\alpha + \beta$ are the roots (or) zeroes of the given polynomial, so

(i) $\alpha – \beta$

The two zeroes of the polynomials are-

$\frac{-b + \sqrt{b^{2} – 4ac}}{2a} – \left ( \frac{-b – \sqrt{b^{2} – 4ac }}{2a} \right )$

= $\frac{-b + \sqrt{b^{2} – 4ac} + b + \sqrt{b^{2} – 4ac}}{2a}$

= $\frac{2\sqrt{b^{2} – 4ac}}{2a}$ = $\frac{\sqrt{b^{2} – 4ac}}{a}$

(ii) $\frac{1}{\alpha } – \frac{1}{\beta }$

= $\frac{\beta – \alpha }{\alpha \beta } = \frac{-\left ( \alpha – \beta \right )}{\alpha \beta }$        ….. E.1

From previous question we know that,

$\alpha – \beta$ = $\frac{\sqrt{b^{2} – 4ac}}{a}$

Also,

$\alpha \beta$ = $\frac{c}{a}$

Putting the values in E.1, we will get

$-\left ( \frac{\frac{\sqrt{b^{2} – 4ac}}{a}}{\frac{c}{a}} \right )$

= $-\left ( \frac{\sqrt{b^{2} – 4ac}}{c} \right )$

(iii) $\frac{1}{\alpha } + \frac{1}{\beta } – 2\alpha \beta$

= $\left [\frac{1}{\alpha } + \frac{1}{\beta } \right ] – 2\alpha \beta$

= $\left [\frac{\alpha + \beta } {\alpha \beta } \right ] – 2\alpha \beta$ …….. E-1

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it in E-1, we will get

$\frac{-b}{a} \times \frac{a}{c} – 2\frac{c}{a}$

= $-\frac{b}{c} – 2 \frac{c}{a}$

= $\frac{-ab – 2c^{2}}{ac}$

= $-\left [ \frac{b}{c} + \frac{2c}{a} \right ]$

(iv) $\alpha ^{2} \beta + \alpha \beta ^{2}$

= $\alpha \beta \left ( \alpha + \beta \right )$ …….. E-1.

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it in E-1, we will get

$\frac{c}{a} \left ( \frac{-b}{a} \right )$

= $\frac{-bc}{ a^{2}}$

(v) $\alpha ^{4} + \beta ^{4}$

= $\left ( \alpha ^{2} + \beta ^{2} \right ) ^{2} – 2\alpha ^{2} \beta ^{2}$

= $\left (\left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ) ^{2} – \left (2\alpha \beta \right )^{2}$ ……. E- 1

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it in E-1, we will get

$\left [ \left (- \frac{b}{a} \right ) – 2 \left ( \frac{c}{a} \right ) \right ]^{2} – \left [ 2 \left (\frac{c}{a} \right )^{2} \right ]$

= $\left [ \frac{b^{2} – 2ac}{a^{2}} \right ]^{2} – \frac{2c ^{2}}{a ^{2}}$

= $\frac{\left ( b^{2} – 2ac \right ) ^{2} – 2a^{2} c^{2}}{a^{4}}$

(vi) $\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}$

= $\frac{a\beta + b + a\alpha + b}{\left ( a\alpha + b \right )\left ( a\beta + b \right )}$

= $\frac{a\left ( \alpha + \beta \right ) + 2b}{a^{2} \alpha \beta + a b \alpha + a b \beta + b^{2}}$

= $\frac{a\left ( \alpha + \beta \right ) + 2b}{a^{2} \alpha \beta + a b \left (\alpha + \beta \right ) + b^{2}}$

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it , we will get

$\frac{b}{ac – b ^{2} + b ^{2}}$

= $\frac{b}{ac }$

(vii) $\frac{\beta }{a\alpha + b} + \frac{\alpha }{a\beta + b}$

= $\frac{\beta \left ( a\beta + b \right ) + \alpha \left ( a\alpha + b \right )}{\left ( a\alpha + b \right )\left ( a\beta + b \right )}$

= $\frac{a \beta ^{2} + b \beta + \alpha a^{2} + b \alpha }{a ^{2} \alpha \beta + a b \alpha + a b \beta + b^{2}}$

= $\frac{a \alpha ^{2} + b \beta ^{2} + b\alpha + b \beta }{a^{2} \times \frac{c}{a} + a b \left ( \alpha + \beta \right ) + b ^{2}}$

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it , we will get

$\frac{ a \left [ \left ( \alpha + \beta \right ) ^{2} + b \left ( \alpha + \beta \right ) \right ]}{ac}$

= $\frac{a\left [ \left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ] – \frac{b ^{2}}{a}}{ac}$

= $\frac{a\left [ \frac{b ^{2}}{a} – \frac{2c}{a} \right ] – \frac{b ^{2}}{a}}{ac}$

= $\frac{a\left [ \frac{b ^{2} – 2c}{a} \right ] – \frac{b ^{2}}{a}}{ac}$

= $\frac{a\left [ \frac{b ^{2} – 2c – b ^{2}}{a} \right ]}{ac}$

= $\frac{b ^{2} – 2c – b ^{2} }{ac}$

= $\frac{ – 2c }{ac}$ = $\frac{ – 2 }{a}$

(viii) $a\left [ \frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha } \right ] + b\left [ \frac{\alpha }{a} + \frac{\beta }{a} \right ]$

= $a \left [ \frac{ \alpha ^{2} + \beta ^{2}}{ \alpha \beta } \right ] + b \left ( \frac{ \alpha ^{2} + \beta ^{2}}{ \alpha \beta } \right )$

= $\frac{ a \left [ \left ( \alpha + \beta \right ) ^{2} – 2 \alpha \beta \right ] + b \left ( \left ( \alpha + \beta \right )^{2} – 2 \alpha \beta \right )}{\alpha \beta }$

Since,

Sum of the zeroes of polynomial =  $\alpha + \beta$ = $\frac{-b}{a}$

Product of zeroes of polynomial =  $\alpha \beta$ = $\frac{c}{a}$

After substituting it , we will get

$\frac{ a \left [ \left ( \frac{-b}{a} \right ) ^{2} – 3 \times \frac{c}{a} \right ] + b \left ( \left ( \frac{-b}{a} \right )^{2} – 2 \frac{c}{a} \right )}{\frac{c}{a} }$

= $\frac{a ^{2}}{c} \left [ \frac{-b ^{2}}{a ^{2}} + \frac{3bc}{a^{2}} + \frac{ b ^{2}}{ a ^{2}} – \frac{2bc}{a ^{2}} \right ]$

= $\left [ \frac{-b ^{2} a ^{2}}{a ^{2} c} + \frac{3bc a ^{2}}{a^{2} c} + \frac{ b ^{2} a^{2}}{ a ^{2}c } – \frac{2bc a ^{2}}{a ^{2} c} \right ]$

= $\frac{- b ^{2}}{ac} + 3b + \frac{b ^{2}}{ac} – 2b$ = b

#### Practise This Question

If x = a cos θ,y=b sin θ,then d3ydx3 is equal to