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

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

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

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

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

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

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

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

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

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