Question

The value of '$3a$' for which one root of the quadratic equation
$(a^2-5a+3)x^2-3(a-1)x+2=0$ is twice as large as other is

Answer 1

Given equation
$(a^2-5a+3)x^2-3(a-1)x+2=0$
For quadratic
$a^2-5a+3\ne 0\Rightarrow a\ne \frac{5\pm \sqrt{13}}{2}$
Let the roots of the equation $m,n$
Now $m=2n$
So,
$m+n=\frac{3(a-1)}{a^2-5a+3}\Rightarrow n=\frac{a-1}{a^2-5a+3}\cdots \left(1\right)$
and
$2n^2=\frac{2}{a^2-5a+3}\Rightarrow n^2=\frac{1}{a^2-5a+3}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
${\left(\frac{a-1}{a^2-5a+3}\right)}^2=\frac{1}{a^2-5a+3}\Rightarrow (a-1{)}^2=(a^2-5a+3)\Rightarrow a^2+1-2a=a^2-5a+3\therefore 3a=2$