Question

Alpha and Beta are the roots of x2+x-2=0,find the value of 1/a2+1/B2

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: \alpha , \: \beta \: are \: roots \: of \: {x}^{2} + x - 2 = 0.$

We know that

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

$\bf\implies \: \alpha + \beta = - \dfrac{1}{1} = - 1$

And

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

$\bf\implies \: \alpha \beta = \dfrac{ - 2}{1} = - 2$

Now, Consider

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

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

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

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

$\rm \:  =  \: \dfrac{ {( - 1)}^{2} - 2( - 2)}{ {( - 2)}^{2} }$

$\rm \:  =  \: \dfrac{1 + 4}{4}$

$\rm \:  =  \: \dfrac{5}{4}$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{1}{ { \alpha }^{2} } + \dfrac{1}{ { \beta }^{2} } = \frac{5}{4} \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Learn More :-

$\boxed{ \tt{ \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \: }}$

$\boxed{ \tt{ \: { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) \: }}$

$\boxed{ \tt{ \: {( \alpha - \beta)}^{2} = {( \alpha + \beta)}^{2} - 4 \alpha \beta }}$