Question

if alpha and beta are the zeries of the polynomial f(x)=3x²-7x-6, find a polynomial whose zeroes are alpha square and beta square​

Answer 1

AnswEr:-

Your answer is 3x²-19x+12.

ExplanaTion:-

Given:-

• $\tt \alpha \:\:And\:\: \beta$ Are zeroes of $\tt f(x) = 3x^2-7x-6.$

To Find:-

• The polynomial whose zeroes are $\tt \alpha^2 \:\:And\:\: \beta^2.$

Concepts Used:-

• Every quadratic polynomial is in the form,

$\tt\longrightarrow {x}^{2} - (sum \: of \: zeroes)x + (product \: of \: zeroes).$

• In a quadratic polynomial if alpha and beta are zeroes then,

$\tt\longrightarrow \alpha + \beta = - \dfrac{b}{a} . \\ \\ \rm \: and \\ \\ \tt\longrightarrow \alpha \beta = \dfrac{c}{a} .$

Where,

• a = Coefficient of x².
• b = Coefficient of x.
• c = Constant Term.

ID Used:-

$\tt\longrightarrow { a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab.$

So Here,

$\tt\longrightarrow \alpha + \beta = - \dfrac{( - 7)}{3} . \\ \\\tt\longrightarrow \alpha + \beta = \dfrac{7}{3}. ..(1)\\ \\ \rm \: and \\ \\ \tt\longrightarrow \alpha \beta = \cancel \dfrac{ - 6}{3} \\ \\ \tt\longrightarrow \alpha \beta = - 2. ..(2)$

Therefore,

$\tt\mapsto { \alpha }^{2} + { \beta }^{2} = ( \alpha + \beta ) {}^{2} - 2 \alpha \beta . \\ \\ \tt\mapsto { \alpha }^{2} + { \beta }^{2} = ( \dfrac{7}{3} ) {}^{2} - 2( - 2). \\ \\ \rm[from \: (1) \: and \: (2)]. \\ \\ \tt\mapsto { \alpha }^{2} + { \beta }^{2} = \dfrac{7}{3} + 4. \\ \\ \tt\mapsto { \alpha }^{2} + \beta {}^{2} = \dfrac{19}{3} ...(3) \\ \\ \rm [by \: taking \: lcm].$

Similarly,

$\tt\mapsto { \alpha }^{2} { \beta }^{2} = ( \alpha \beta ) {}^{2}. \\ \\ \tt\mapsto { \alpha }^{2} \beta {}^{2} = ( - 2) {}^{2} \\ \\ \tt\mapsto { \alpha }^{2} { \beta }^{2} = 4...(4)$

So the sum of product of the zeroes of polynomial having zereos alpha² and beta² is 19/3 and 4 respectively.

So the required polynomial is:-

$\tt\leadsto {x}^{2} - ( { \alpha }^{2} + { \beta }^{2} )x + ( \alpha {}^{2} \beta {}^{2}). \\ \\ \tt = {x}^{2} - ( \dfrac{19}{3} )x + (4). \\ \\ \tt[from \: \: (3) \: \: and \: \: (4)]. \\ \\ \tt = {x}^{2} - \dfrac{19}{3}x + 4 \\ \\ \tt = 3 {x}^{2} - 19x + 12.$

Therefore the required polynomial is 3x²-19x+12.

$\rule{200}2$

Now you may have a doubt why the polynomial is,

$\tt\rightarrow 3x^2-19x+12 \:and\:not, \\ \\ \dfrac{3x^2-19x+12}{3}$

This is because when we will find out the zeroes of this polynomial it must be equal to zero and than it will solve like,

$\tt\rightarrow \dfrac{3x^2-19x+12}{3}=0. \\ \\ \tt\rightarrow 3x^2-19x+12 = 0\times3.\\ \\ \tt\rightarrow 3x^2-19x+12 = 0.$

That is why we had directly written it as $\tt 3x^2-19x+12$ Instead of $\tt\dfrac{3x^2-19x+12}{3}$.

$\rule{200}2$