Question

If $\sqrt{{3x}^2-7x-30}+\sqrt{{2x}^2-7x-5}=x+5$ , then $x$ is equal to

• A. $2$
• B. $3$
• C. $6$
• D. $5$

Answer 1

Answer: (C)

$6$

$\sqrt{{3x}^2-7x-30}+\sqrt{{2x}^2-7x-5}=x+5$

$\Rightarrow$ $\sqrt{3x^2-7x-30}=(x+5)-\sqrt{2x^2-7x-5}$

$\Rightarrow$ $3x^2-7x-30=x^2+10x+25-2\sqrt{2x^2-7x-5}(x+5)+2x^2-7x-5$ [ Squaring both sides ]

$\Rightarrow$ $2(x+5)\sqrt{2x^2-7x-5}=10x+50$

$\Rightarrow$ $2(x+5)\sqrt{2x^2-7x-5}=10(x+5)$

$\Rightarrow$ $\sqrt{2x^2-7x-5}=5$

$\Rightarrow$ $2x^2-7x-5=25$ [ Squaring both sides ]

$\Rightarrow$ $2x^2-7x-30=0$

$\Rightarrow$ $2x^2-12x+5x-30=0$

$\Rightarrow$ $2x(x-6)+5(x-6)=0$

$\Rightarrow$ $(x-6)(2x+5)=0$

$\therefore$ $x=6$ or $x=\frac{-5}{2}$