Question

if the zeroes of polynomial ax^2+x+a are equal then value of a is
(a)+1/2 (b)-1 (c)+1/3 (d) none of these​

Answer 1

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

Given quadratic polynomial is

$\rm :\longmapsto\:f(x) = {ax}^{2} + x + a \: have \: equal \: zeroes$

Let assume that

$\rm :\longmapsto\:f(x) = {ax}^{2} + x + a \: have\: zeroes \: \alpha , \alpha$

We know

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

So,

$\rm \implies\: \alpha + \alpha = - \: \dfrac{1}{a}$

$\rm \implies\: 2\alpha = - \: \dfrac{1}{a}$

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

Also, we know that

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

$\rm \implies\: \alpha \times \alpha = \dfrac{a}{a}$

$\rm \implies\: { \alpha }^{2} = \dfrac{a}{a}$

$\rm \implies\: {\bigg[\dfrac{1}{2a} \bigg]}^{2} = 1$

$\rm :\longmapsto\:\dfrac{1}{2a} = \: \pm \: 1$

$\bf\implies \:a \: = \: \pm \: \dfrac{1}{2}$

$\bf\implies \:a \: = \: \dfrac{1}{2} \: \: or \: \: - \: \dfrac{1}{2}$

So,

• Option (d) is correct

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Additional Information :-

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

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

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

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