Question

Consider the equation $\sqrt{3x^2-8x+1}+\sqrt{9x^2-24x-8}=3.$ It is known that the largest root of the equation is $-k$ times the smallest root. The value of $k$ is
(correct answer + 3, wrong answer 0)

Answer 1

Let $y=\sqrt{3x^2-8x+1}$
Then the equation becomes
$y+\sqrt{3y^2-11}=3$
Then $3y^2-11=(3-y{)}^2$
$\Rightarrow 3y^2-11=9+y^2-6y$
$\Rightarrow y^2+3y-10=0$
$\Rightarrow y=2$ or $y=-5(\because y\ge 0)$

$\therefore \sqrt{3x^2-8x+1}=2$
$\Rightarrow 3x^2-8x+1=4$
$\Rightarrow 3x^2-8x-3=0$
$\Rightarrow x=3$ and $x=\frac{-1}{3}$
So, $3=(-k)\left(\frac{-1}{3}\right)$
$\Rightarrow k=9$