Mathematics - Quadratic Equation Question with Solution | TestHub

$\left(a^2-5a+3a\right)x^2+(3a-1)x+2=0$ $\_\_\_\_$ Let $\alpha$ and $2\alpha$ be the roots of (1), then $\left(a^2-5a+3\right){\alpha }^2+(3a-1)\alpha +2=0$ $\text{ and }\left(a^2-5a+3\right)\left(4{\alpha }^2\right)+(3a-1)(2\alpha )+2=0$ Multiplying (2) by 4 and subtracting it form (3) we get $(3a-1)(2\alpha )+6=0$ Clearly $a\ne 1/3$. Therefore, $\alpha =-3/(3a-1)$ Putting this value in (2) we get $\begin{array}{rlrl}& \left(a^2-5a+3\right)(9)-(3a-1{)}^2(3)+2(3a-1{)}^2=0& & \end{array}$

$$\Rightarrow 9a^2-45a+27-\left(9a^2-6a+1\right)=0\Rightarrow -39a+26=0\Rightarrow a=2/3$$ For $x=2/3$, the equation becomes $x^2+9x+18=0$, whose roots are $-3,-6$.