Question

If the roots of the equation
${ax}^{2} - {bx} + 5c = 0$
are in the ratio 4:5 , then

A)
$ab = 18 {c}^{2}$
B)
$81 {b}^{2} = 4ac$
C)
$bc = {a}^{2}$
D)
$4 {b}^{2} = 81ac$



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Answer 1

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

Given that

$\rm :\longmapsto\:Roots \: of \: {ax}^{2} - bx + 5c = 0 \: are \: in \: the \: ratio \: 4: 5.$

So,

$\rm :\longmapsto\:Let \: assume \:that \: roots \: be \: 4 \alpha \: and \: 5 \alpha$

We know that,

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

$\rm :\longmapsto\:4 \alpha + 5 \alpha = - \dfrac{( - b)}{a}$

$\rm :\longmapsto\:9\alpha = \dfrac{b}{a}$

$\red{\rm :\longmapsto\:\alpha = \dfrac{b}{9a} - - - (1)}$

Also,

We know that,

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

$\rm :\longmapsto\:4 \alpha \times \cancel 5 \alpha = \dfrac{ \cancel{5} \: c}{a}$

$\rm :\longmapsto\:4 { \alpha }^{2} = \dfrac{c}{a}$

On substituting the value from equation (1), we get

$\rm :\longmapsto\:4 {\bigg(\dfrac{b}{9a} \bigg) }^{2} = \dfrac{c}{a}$

$\rm :\longmapsto\:\dfrac{4 {b}^{2} }{81 {a}^{2} } = \dfrac{c}{a}$

$\rm :\longmapsto\:\dfrac{4 {b}^{2} }{81 {a}} = c$

$\purple{\bf\implies \: {4b}^{2} = 81ac}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \purple{\underbrace{\boxed{ \purple{ \bf{ \: Option \: (d) \: is \: correct.}}}}}$

Additional Information :-

Nature of roots :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.

• If Discriminant, D > 0, then roots of the equation are real and unequal.

• If Discriminant, D = 0, then roots of the equation are real and equal.

• If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.

Where,

• Discriminant, D = b² - 4ac