Question

Q25. If α and β are the zeroes of the quadratic polynomial 2x²+5x+1, then find the value of α²+ β².​

Answer 1

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

Given that,

$\rm :\longmapsto\: \alpha , \beta \: are \: zeroes \: of \: polynomial \: {2x}^{2} + 5x + 1$

We know

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

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

And

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

$\bf\implies \: \alpha + \beta = - \: \dfrac{5}{2}$

Now, Consider

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

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

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

$\rm \: = \: {\bigg[ - \dfrac{5}{2} \bigg]}^{2} - 2 \times \dfrac{1}{2}$

$\rm \: = \: \dfrac{25}{4} - 1$

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

$\rm \: = \: \dfrac{21}{4}$

Hence,

$\red{\rm :\longmapsto\: \boxed{ \tt{ \: { \alpha }^{2} + { \beta }^{2} \: } = \frac{21}{4} \: }}$

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Additional Information :-

$\red{\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then}$

$\boxed{ \bf{ \: \alpha + \beta + \gamma = - \dfrac{b}{a}}}$

$\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \dfrac{c}{a}}}$

$\boxed{ \bf{ \: \alpha \beta \gamma = - \dfrac{d}{a}}}$