Question

f the roots of ax2 + bx + c = 0 are in the ratio m : n, then
(a) mna² = (m + n) c²
(b) mnb² = (m + n) ac
(c) mn b² = (m + n)² ac
(d) mnb² = (m – n)² ac​

Answer 1

Answer

The correct option is :

(c) mnb² = (m + n)²ac

$\bf \large \underline{Given : }$

The quadratic equation is :

• ax² + bx + c = 0
• The roots of the equation are in the ratio m:n

$\bf \large \underline{Solution : }$

Let us consider α and β are the zeroes of the polynomial ax² + bx + c

$\sf \alpha : \beta = m:n \\\\ \sf\implies \dfrac{\alpha}{\beta} = \dfrac{m}{n}$

We have from relation of sum of roots ,

$\sf sum \: of \: the \: zeroes \: = \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\sf\implies \alpha + \beta = \dfrac{-b}{a} \\\\ \sf\implies (\alpha+\beta)^{2} =( \dfrac{-b}{a})^{2} \\\\ \sf\implies \alpha^{2} + \beta^{2} + 2\alpha\beta = \dfrac{b^{2}}{a^{2}} \longrightarrow (1)$

Again, from relation of product of roots ,

$\sf product \: of \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\sf\implies \alpha\beta = \dfrac{c}{a} \longrightarrow (2)$

Dividing (1) by (2) we have ,

$\sf \implies \dfrac{\alpha^{2}+ \beta^{2} + 2\alpha\beta}{\alpha\beta} = \dfrac{\dfrac{b^{2}}{a^{2}}}{\dfrac{c}{a}} \\\\ \sf\implies \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} + 2 = \dfrac{b^{2}}{ac} \\\\ \sf\implies \dfrac{m}{n} + \dfrac{n}{m} + 2 = \dfrac{b^{2}}{ac} \\\\ \sf\implies \dfrac{m^{2} + n^{2} + 2mn}{mn} = \dfrac{b^{2}}{ac} \\\\ \sf\implies \dfrac{(m+n)^{2}}{mn} = \dfrac{b^{2}}{ac} \\\\ \sf\implies (m+n)^{2}ac = mnb^{2} \\\\ \sf\implies mnb^{2} = (m+n)^{2} ac$