Question

Find the value of 'q' so that the equation $2x^2-3qx+5q=0$ has one root which is twice the other.

Answer 1

Given : $2x^2-3qx+5q=0$ ...... $\left(i\right)$

Standard form of quadratic equation is $a{x}^2+bx+c=0$

From $\left(i\right)$

$a=2,b=-3q,c=5q$

Let $m,n$ be roots of $\left(i\right)$

We know that $m+n=-\frac{m}{n}$ and $m\cdot n=\frac{c}{a}$

Since, one root is twice the other $⟹n=2m$

$\therefore m+2m=-\frac{(-3q)}{2}$ and $m\cdot 2m=\frac{c}{a}$

$⟹3m=\frac{3q}{2}$ and $2m^2=\frac{5q}{2}$

$⟹m=\frac{q}{2}$ and $4m^2-5q=0$

So, $4\times \frac{q^2}{4}-5q=0$

$⟹q^2-5q=0$

$⟹q(q-5)=0$

$⟹q=5$