# Thread: [SOLVED] How to solve this question?

1. ## [SOLVED] How to solve this question?

the roots of the quadratic equation $x^2 + mx + n = 0$ are $\alpha$ and $\beta$. If the roots are in the ratio a:b, show that m²ab = n(a + b)²

2. Hello, Mark!

The roots of the quadratic equation $x^2 + mx + n \:=\: 0$ are $\alpha$ and $\beta$.

If the roots are in the ratio $a:b$, show that: . $m^2ab \:=\: n(a + b)^2$

We have: . $\begin{Bmatrix}\alpha &=& ak \\ \beta &=& bk \end{Bmatrix}\;\; \text{ for some constant }k.$

And we have: . $\begin{array}{cccccccc}
\alpha + \beta &=& \text{-}m & \Longrightarrow & ak + bk &=& \text{-}m & {\color{blue}[1]} \\ \alpha\beta &=& n & \Longrightarrow & ak\cdot bk &=& n & {\color{blue}[2]} \end{array}$

From [1], we have: . $(a+b)k \:=\:\text{-}m \quad\Rightarrow\quad k \:=\:\frac{\text{-}m}{a+b}\;\;{\color{blue}[3]}$

From [2], we have: . $abk^2 \:=\:n\;\;{\color{blue}[4]}$

Substitute [3] into [4]: . $ab\left(\frac{\text{-}m}{a+b}\right)^2 \:=\:n \quad\Rightarrow\quad ab\,\frac{m^2}{(a+b)^2} \:=\:n$

. . Therefore: . $m^2ab \;=\;n(a+b)^2$