Question

Solve the equation $\sqrt{4x^2-7x-15}-\sqrt{x^2-3x}=\sqrt{x^2-9}$

• A. $2,3$
• B. $1,6$
• C. $-1,3$
• D. $1,3$

Answer 1

Answer: (D)

$1,3$

Given equation is $\sqrt{4x^2-7x-15}=\sqrt{x^2-9}+\sqrt{x^2-3x}$

$\Rightarrow \sqrt{x-3}\left(\sqrt{4x+5}\right)=\sqrt{x-3}(\sqrt{x+3}+\sqrt{x})$

Therefore $x=3$ is one solution and $\sqrt{4x+5}=\sqrt{x+3}+\sqrt{x}$

By squaring above equation on both sides , we get $x+1=\sqrt{x(x+3)}$

Again square it on both sides , we get $x^2+2x+1=x^2+3x$

$\Rightarrow x=1$

Therefore option $D$ is correct