Question

If alpha and beta are the zeroes of the quadratic polynomial x^2 + 2 + 1 , then find the quadratic polynomial whose zeroes are alpha square beta and beta square alpha.

Answer 1

$\mathfrak{\large{\underline{\underline{Given:-}}}}$

$\alpha$and $\beta$ are the zeroes of the quadratic equation.

$\mathfrak{\large{\underline{\underline{To find:-}}}}$

The quadratic equation whose zeroes is $\ { \alpha }^{2} + { \beta }^{2}$.

$\mathfrak{\large{\underline{\underline{Solution:-}}}}$

By comparing the quadratic equation with its standard form.

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Let a = 1 ,b = 2 and c = 1

Then,

$\boxed{\sf{ \alpha + \beta = \frac{ - b}{a} }}$

$\implies$ $\ \alpha + \beta = \frac{ - 2}{1}$

$\implies$ $\ \alpha + \beta = -2$ eq. 1.

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Now,

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

$\implies$ $\ alpha \beta = \frac{1}{1}$

$\implies$ $\alpha \beta = 1$ eq. 2.

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By using suitable identities

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

from eq.1. and eq. 2. we get,

$\implies$ $\{( -2 )}^{2} = { \alpha }^{2} + { \beta }^{2} + 2 \times 1$

$\implies$ $\{(4 - 1)} = { \alpha }^{2} + { \beta }^{2}$

$\implies$ $\ { \alpha }^{2} + { \beta }^{2} = 3$

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The required quadratic equation is given by :-

$\boxed{\sf{{x}^{2} - ( \alpha^2 + \beta^2 )x + \alpha^2 \beta^2 }}$

$\implies$ $\ {x}^{2} - 3x + {1}^{2}$

$\implies$ $\ {x}^{2} - 3x + 1$ is required quadratic equation.