# How do you solve sqrt(x+5) + sqrt(x+15) = sqrt(9x+40)?

Jul 25, 2015

Square, rearrange and square again to get a quadratic, one of whose roots is a valid solution:

$x = \frac{- 20 + 2 \sqrt{55}}{9} \cong - 0.574$

#### Explanation:

Note that both sides are non-negative, being the sums of non-negative square roots. So squaring both sides will not introduce spurious solutions.

So square both sides to get:

$9 x + 40 = \left(x + 5\right) + 2 \sqrt{x + 5} \sqrt{x + 15} + \left(x + 15\right)$

$= 2 x + 20 + 2 \sqrt{{x}^{2} + 20 x + 75}$

Subtract $2 x + 20$ from both ends to get:

$7 x + 20 = 2 \sqrt{{x}^{2} + 20 x + 75}$

Note that a valid solution will require $7 x + 20 \ge 0$.

Square both sides (which may introduce extraneous solutions with $7 x + 20 < 0$) to get:

$49 {x}^{2} + 280 x + 400 = 4 {x}^{2} + 80 x + 300$

Subtract the right hand side from the left to get:

$45 {x}^{2} + 200 x + 100 = 0$

Divide through by $5$ to get:

$9 {x}^{2} + 40 x + 20 = 0$

Solve using the quadratic formula:

$x = \frac{- 40 \pm \sqrt{{40}^{2} - \left(4 \times 9 \times 20\right)}}{2 \cdot 9}$

$= \frac{- 40 \pm \sqrt{1600 - 720}}{18}$

$= \frac{- 40 \pm \sqrt{880}}{18}$

$= \frac{- 40 \pm \sqrt{16 \cdot 55}}{18}$

$= \frac{- 40 \pm 4 \sqrt{55}}{18}$

$= \frac{- 20 \pm 2 \sqrt{55}}{9}$

$\frac{- 20 - 2 \sqrt{55}}{9} \cong - 3.870$

$7 \left(- 3.870\right) + 20 < 0$ so this solution is spurious.

$\frac{- 20 + 2 \sqrt{55}}{9} \cong - 0.574$

$7 \left(- 0.574\right) + 20 > 0$ so this solution is valid.