Question

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

Answer 1

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

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

${\sqrt{3x^2-7x-30}}^2={x-5+\sqrt{2x^2-7x-5}}^2$ by squaring both sides

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

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

$\Rightarrow 3x^2-7x-30=3x^2-17x+20+2(x-5)\sqrt{(2x^2-7x-5)}$

$\Rightarrow -7x+17x-30-20=2(x-5)\sqrt{(2x^2-7x-5)}$

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

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

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

$\Rightarrow 25=2x^2-7x-5$ by squaring on both sides

$\Rightarrow 2x^2-7x-5-25=0$

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

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

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

For $x=\frac{-5}{2}$ the $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$ equation becomes

$\sqrt{3{\left(\frac{-5}{2}\right)}^2-7\times \frac{-5}{2}-30}-\sqrt{2{\left(\frac{-5}{2}\right)}^2-7\times \frac{-5}{2}-5}=\frac{-5}{2}-5$

$=\sqrt{\frac{75}{4}+\frac{35}{2}-30}-\sqrt{2\times \frac{25}{4}+\frac{35}{2}-5}$

$=\sqrt{\frac{145-120}{4}}-\sqrt{\frac{40}{2}}$

$=\sqrt{\frac{25}{4}}-\sqrt{20}=\frac{5}{2}-4\sqrt{5}\ne$ RHS$x-5=\frac{-5}{2}-5=\frac{-15}{2}$

Hence $x=\frac{-5}{2}$ is not a solution

For $x=6$ the $\sqrt{3x^2-7x-30}-\sqrt{2x^2-7x-5}=x-5$ equation becomes

$\sqrt{3{\left(6\right)}^2-7\times 6-30}-\sqrt{2{\left(6\right)}^2-7\times 6-5}=6-5$

$=\sqrt{108-72}-\sqrt{72-47}=1$

$=\sqrt{36}-\sqrt{25}=6-5=1$

Thus, $x=6$ is the only solution for the above equation