Question

the condition that one root of ax2+bx+c=0 may be square the other is​

Answer 1

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

We know,

$\sf \: If \: \alpha , \beta \: are \: the \: zeroes \: of \: {ax}^{2} + bx + c = 0\: then$

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

and

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

Let's solve the problem now!!

Given that

• One zero of ax² + bx + c = 0 is square of other

$\sf \: Let \:y\:and \:{y}^{2} \: be \: the \: zeroes\:of\:{ax}^{2}+bx + c = 0\:$

So,

$\rm :\longmapsto\:{\purple{\sf Product\ of\ the\ zeroes=\dfrac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:y \times {y}^{2} = \dfrac{c}{a}$

$\bf\implies \: {y}^{3} = \dfrac{c}{a}$

Also,

$\rm :\longmapsto\:{\purple{\sf Sum\ of\ the\ zeroes=\dfrac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:y + {y}^{2} = - \dfrac{b}{a}$

On Cubing both sides, we get

$\rm :\longmapsto\: {(y + {y}^{2})}^{3} = - \dfrac{ {b}^{3} }{ {a}^{3} }$

$\rm :\longmapsto\: {y}^{3} + {y}^{6} + 3(y)( {y}^{2})(y +{y}^{2})= - \dfrac{ {b}^{3} }{ {a}^{3} }$

$\rm :\longmapsto\: {y}^{3} + {y}^{6} + 3{y}^{3}(y +{y}^{2})= - \dfrac{ {b}^{3} }{ {a}^{3} }$

$\rm :\longmapsto\:\dfrac{c}{a} + {\bigg(\dfrac{c}{a}\bigg) }^{2} + 3\bigg(\dfrac{c}{a}\bigg)\bigg( - \dfrac{b}{a}\bigg) = -\dfrac{ {b}^{3} }{ {a}^{3} }$

$\: \: \: \: \: \: \: \: \boxed{ \because \purple{ \sf \: {y}^{3} = \dfrac{c}{a} \: \: \: \: and \: \: \: \: y + {y}^{2} = - \dfrac{b}{a}}}$

$\rm :\longmapsto\:\dfrac{c}{a} + \dfrac{ {c}^{2} }{ {a}^{2} } - 3\dfrac{bc}{ {a}^{2} } = - \dfrac{ {b}^{3} }{ {a}^{3} }$

$\rm :\longmapsto\:\dfrac{ac + {c}^{2} - 3bc }{ {a}^{2} } = - \dfrac{ {b}^{3} }{ {a}^{3} }$

$\rm :\longmapsto\: {ca}^{2} + {ac}^{2} - 3abc = - {b}^{3}$

$\rm :\longmapsto\: {ca}^{2} + {ac}^{2} - 3abc + {b}^{3}=0$

Hence, the required condition is

$\: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \purple{ \bf\: {ca}^{2} + {ac}^{2}+ {b}^{3} - 3abc=0}}}$

Additional Information :-

Important Identities :-

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

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