Thank you for reporting, we will resolve it shortly
Q.
If $\tan ^{2} \frac{3 \pi}{8}$ is a root of the equation $2 x ^{2}-3 ax +4 b =0$, where $a, b \in Q$, then the value of $a+4 b$ can be equal to:
Complex Numbers and Quadratic Equations
Solution:
$\tan ^{2} \frac{3 \pi}{8}=(\sqrt{2}+1)^{2}=3+2 \sqrt{2}$
$\therefore x=3+2 \sqrt{2}$ satisfy the equation
$2 x^{2}-3 a x+4 b=0... $(1)
$(x-3)^{2}=8 \Rightarrow x^{2}-6 x+1=0...$(2)
comparing (1) and (2), we get
$\therefore a =4$ and $ b =1 / 2 $
$\Rightarrow a +4 b =6$