Question

of alpha and beta are zeros of the polynomial 2x^2-5x+3 find alpha^2 + beta^2

Answer 1

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$\huge\red{\mathfrak{Solution}}$
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$2 {x}^{2} - 5x + 3$

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

$a = 2 \\ b = - 5 \\ c = 3$

$= > \alpha + \beta = \frac{ - ( - 5)}{2} \\ = > \alpha + \beta = \frac{5}{2}$

$\alpha \beta = \frac{c}{a} \\ = > \alpha \beta = \frac{3}{2}$

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

$putting \: the \: value \: of \: \alpha \beta \: nd \: ( \alpha + \beta ) \\ in \: above \: formulae$

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

$=>\frac{25}{4} = { \alpha }^{2} + { \beta }^{2} + \frac{6}{2}$

$=>( \frac{25}{4} - \frac{6}{2} ) = { \alpha }^{2} + { \beta }^{2}$

$=>( \frac{25 - 12}{4} ) = { \alpha }^{2} + { \beta }^{2}$

${\green{\boxed{=> { \alpha }^{2} + { \beta }^{2} = \frac{13}{4} }}}$

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