Question

Solve the following equations:
$\sqrt{3x^2-2x+9}+\sqrt{3x^2-2x-4}=13$.

Answer 1

$\sqrt{3x^2-2x+9}+\sqrt{3x^2-2x-4}=13$

Put $3x^2-2x=t$

$\sqrt{t+9}+\sqrt{t-4}=13\sqrt{t+9}=13-\sqrt{t-4}$

squaring both sides

$t+9=169+t-4-26\sqrt{t-4}-156=-26\sqrt{t-4}\sqrt{t-4}=6$

Again squaring both sides

$t-4=36t=40\Rightarrow t=403x^2-2x=t3x^2-2x=40{3x}^2-2x-40=0{3x}^2-12x+10x-40=03x(x-4)+10(x-4)=0(3x+10)(x-4)=0x=-\frac{10}{3},4$