# Solve the equation?

## $\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6} = \sqrt{6 {x}^{2} - 5 x - 39}$

Jun 25, 2018

$x = 3$

#### Explanation:

$\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6} = \sqrt{6 {x}^{2} - 5 x - 39}$

${\left(\sqrt{{x}^{2} + 4 x - 21} + \sqrt{{x}^{2} - x - 6}\right)}^{2} = {\left(\sqrt{6 {x}^{2} - 5 x - 39}\right)}^{2}$

${x}^{2} + 4 x - 21 + 2 \cdot \sqrt{{x}^{2} + 4 x - 21} \cdot \sqrt{{x}^{2} - x - 6} + {x}^{2} - x - 6 = 6 {x}^{2} - 5 x - 39$

$2 {x}^{2} + 3 x - 27 + 2 \sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 6 {x}^{2} - 5 x - 39$

$2 \sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 4 {x}^{2} - 8 x - 12$

$\sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)} = 2 {x}^{2} - 4 x - 6$

${\left(\sqrt{\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right)}\right)}^{2} = {\left(2 {x}^{2} - 4 x - 6\right)}^{2}$

$\left({x}^{2} + 4 x - 21\right) \left({x}^{2} - x - 6\right) = {\left(2\right)}^{2} {\left({x}^{2} - 2 x - 3\right)}^{2}$

$\left(x + 7\right) \left(x - 3\right) \left(x - 3\right) \left(x + 2\right) = 4 {\left(\left(x - 3\right) \left(x + 1\right)\right)}^{2}$

$\left(x + 7\right) {\left(x - 3\right)}^{2} \left(x + 2\right) = 4 {\left(x - 3\right)}^{2} {\left(x + 1\right)}^{2}$

${\left(x - 3\right)}^{2} \left(\left(x + 7\right) \left(x + 2\right) - 4 {\left(x + 1\right)}^{2}\right) = 0$

${\left(x - 3\right)}^{2} \left({x}^{2} + 9 x + 14 - 4 \left({x}^{2} + 2 x + 1\right)\right) = 0$

${\left(x - 3\right)}^{2} \left({x}^{2} + 9 x + 14 - 4 {x}^{2} - 8 x - 4\right) = 0$

${\left(x - 3\right)}^{2} \left(- 3 {x}^{2} + x + 10\right) = 0$

$- {\left(x - 3\right)}^{2} \left(3 {x}^{2} - x - 10\right) = 0$

$- {\left(x - 3\right)}^{2} \left(3 x + 5\right) \left(x - 2\right) = 0$

$x = 3$ or $x = - \frac{5}{3}$ or $x = 2$

However, as @Mark D points out, all the solutions but $x = 3$ give negative numbers within the square roots and so only $x = 3$ is valid:

$\sqrt{{x}^{2} - x - 6}$

$\sqrt{{2}^{2} - 2 - 6} \implies \sqrt{4 - 2 - 6} \implies \sqrt{- 4}$

$\sqrt{{\left(- \frac{5}{3}\right)}^{2} + \frac{5}{3} - 6} \implies \sqrt{\frac{25}{9} + \frac{15}{9} - \frac{54}{9}} \implies \sqrt{- \frac{14}{9}}$