Question

The value of '3a', for which one root of the quadratic polynomial $(a^2-5a+3)x^2+(3a-1)x+2$ is twice as large as the other, is______.

Answer 1

Let the root be $p,2p$

Sum $=3p=\frac{1-3a}{a^2-5a+3}\Rightarrow p=\frac{1}{3}\frac{1-3a}{a^2-5a+3}................\left(1\right)$

Product $=2p^2=\frac{2}{a^2-5a+3}\Rightarrow p^2=\frac{1}{a^2-5a+3}...............\left(2\right)$

From (1) and (2)

$\frac{1}{9}\frac{(1-3a{)}^2}{(a^2-5a+3{)}^2}=\frac{1}{a^2-5a+3}\Rightarrow 1+9a^2-6a=9a^2-45a+27\Rightarrow a=\frac{2}{3}$

Hence, $3a=3.\frac{2}{3}=2$