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 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

Then, the quadratic polynomial is-

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

Hence, the quadratic polynomial is

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

= 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 = \(\frac{1}{2 \alpha + \beta } \times \frac{1}{2\beta + \alpha }\)

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

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

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

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

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

= \(\frac{1}{-10 + 26}\) = \(\frac{1}{16}\)

So, the quadratic polynomial is,

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

= 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

So, quadratic polynomial is:

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

= \(x^{2} – 6x + 11\)

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

= \(\frac{2}{6} = \frac{1}{3}\)

So, the quadratic polynomial is,

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

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

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