Question

If $f\left(x\right)=x^2-3x+1$ and $f\left(2\alpha \right)=2f\left(\alpha \right)$, then find the value of $\alpha$.

Answer 1

$f\left(x\right)=x^2-3x+1$

$f\left(2\alpha \right)=(2\alpha {)}^2-3(2\alpha )+1=4{\alpha }^2-6\alpha +1$

$f\left(2\alpha \right)=2f\left(\alpha \right)$

$\therefore 4{\alpha }^2-6\alpha +1=2({\alpha }^2-3\alpha +1)$

$4{\alpha }^2-6\alpha +1=2{\alpha }^2-b\alpha +2$

$4{\alpha }^2+1=2{\alpha }^2+2$

$2{\alpha }^2=1$ $\alpha =\pm \frac{1}{\sqrt{2}}$