Question

if alpha and beta are the zeroes of the polynomial 2x^2-4x+5,find the values of 1/alpha^2+1/beta^2
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Answer 1

$\large\underline\blue{\bold{Given \:  :-  }}$

$\rm : \implies \: \alpha \: and \: \beta \: are \: zeroes \: of \: polynomial \: {2x}^{2} - 4x + 5$

$\large\underline\blue{\bold{To \: Find - }}$

$\rightarrow \rm \: the \: value \: of \: \dfrac{1}{ { \alpha }^{2} } + \dfrac{1}{ { \beta }^{2} }$

$\large\underline\purple{\bold{Solution :- }}$

Now,

We know that,

$\boxed{\purple{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

OR

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

And

$\boxed{\purple{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

OR

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

So,

$\rightarrow \rm \: \alpha + \beta = \frac{-b}{a}$

$\rightarrow \rm \: \alpha + \beta = \frac{-( - 4)}{2}$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \: \alpha + \beta \: = \: 2}}$

And

$\rightarrow \rm \: \alpha \beta = \frac{c}{a}$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \: \alpha \beta \: = \: \dfrac{5 }{2} }}$

Now, Consider

$\rightarrow \rm \: \dfrac{1}{ { \alpha }^{2} } + \dfrac{1 }{ { \beta }^{2} }$

$\rm : \implies \:\dfrac{ { \beta }^{2} + { \alpha }^{2} }{ { \alpha }^{2} { \beta }^{2} }$

$\rm : \implies \:\dfrac{ { \alpha }^{2} + { \beta }^{2} + 2 \alpha \beta - 2 \alpha \beta }{ {( \alpha \beta )}^{2} }$

$\rm : \implies \:\dfrac{ {( \alpha + \beta )}^{2} - 2 \alpha \beta }{ {( \alpha \beta )}^{2} }$

On substituting the values, we get

$\rm : \implies \:\dfrac{ {(2)}^{2} - 2 \times \dfrac{5}{2} }{ {(\dfrac{5}{2} )}^{2} }$

$\rm : \implies \:(4 -5 ) \times \dfrac{4}{25}$

$\rm : \implies \: - \: \dfrac{4}{25}$

$\bigstar \: \: \boxed{ \pink{ \rm : \implies \:\dfrac{1}{ { \alpha }^{2} } \: + \: \dfrac{1}{ { \beta }^{2} } \: = \: - \: \dfrac{4}{25} \: }}$