Question

Find the condition that the ratio between the roots of the equation ax² + bx + c = 0 may be m : n.
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Answer 1

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

Given that, the ratio between the roots of the equation ax² + bx + c = 0 may be m : n.

Let assume that

$\rm \: m \alpha \: and \: n \alpha \: be \: the \: roots \: of \: equation \: {ax}^{2} + bx + c = 0$

We know,

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

$\rm \: m \alpha + n \alpha = \: - \: \dfrac{b}{a}$

$\rm \: (m + n )\alpha = \: - \: \dfrac{b}{a}$

$\rm\implies \:\alpha = \: - \: \dfrac{b}{a(m + n)} - - - (1)$

Also, we know that

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

So,

$\rm \: m \alpha \times n \alpha = \dfrac{c}{a}$

$\rm \: mn {( \alpha )}^{2} = \dfrac{c}{a}$

On substituting the value from equation (1) we get

$\rm \: mn {\bigg( - \frac{b}{a(m + n)} \bigg)}^{2} = \dfrac{c}{a}$

$\rm \: \dfrac{mn {b}^{2} }{ {a}^{2} {(m + n)}^{2} } = \dfrac{c}{a}$

$\rm\implies \:mn {b}^{2} = ac {(m + n)}^{2}$

$\rule{190pt}{2pt}$

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