Question

The solution set of the equation $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$ is

• A. $\{6\}$
• B. $\left\{6,-\frac{5}{2}\right\}$
• C. $\{5,6\}$
• D. $\left\{5,\frac{7}{2}\right\}$

Answer 1

Answer: (C)

Given equation can be rewritten as $\sqrt{3x^2-7x-30}=(x-5)+\sqrt{2x^2-7x-5}$ On squaring both sides, we get $3x^2-7x-30=x^2+25-10x+(2x^2-7x-5)$ $+2(x-5)\sqrt{2x^2-7x-5}$
$\Rightarrow$ $10x-50=2(x-5)\sqrt{2x^2-7x-5}$
$\Rightarrow$ $(x-5)(5-\sqrt{2x^2-7x-5)}=0$
$\Rightarrow$ either $x-5=0$ or $5-\sqrt{2x^2-7x-5}=0$
$\Rightarrow$ $x=5$ or $5=\sqrt{2x^2-7x-5}$
$\Rightarrow$ $x=5$ or $2x^2-7x-30=0$
$\Rightarrow$ $x=5$ or $x=-5/2,6$ But $x=-\frac{5}{2}$ is not satisfied the given equation. Hence, required roots are $\{5,6\}$ .