# How do you solve sqrt(2x+4)+sqrt(3x-5)=4?

Sep 6, 2017

Given: $\sqrt{2 x + 4} + \sqrt{3 x - 5} = 4 \text{ }$

If we mulitply both sides by the conjugate, $\sqrt{2 x + 4} - \sqrt{3 x - 5}$, it will eliminate the radicals from the left side:

$2 x + 4 - \left(3 x - 5\right) = 4 \left(\sqrt{2 x + 4} - \sqrt{3 x - 5}\right) \text{ }$

Combine like terms and distribute the 4:

$9 - x = 4 \sqrt{2 x + 4} - 4 \sqrt{3 x - 5} \text{ }$

Flip equation :

$4 \sqrt{2 x + 4} - 4 \sqrt{3 x - 5} = 9 - x \text{ }$

Multiply both sides of equation  by 4:

$4 \sqrt{2 x + 4} + 4 \sqrt{3 x - 5} = 16 \text{ }$

We can make the term $4 \sqrt{3 x - 5}$ disappear by adding equation  to equation :

$8 \sqrt{2 x + 4} = 25 - x \text{ }$

Square both sides:

$64 \left(2 x + 4\right) = {x}^{2} - 50 x + 625$

Distribute the 64:

$128 x + 256 = {x}^{2} - 50 x + 625$

Combine like terms:

${x}^{2} - 178 x + 369 = 0$

$x = 89 - 8 \sqrt{118}$ and $x = 89 + 8 \sqrt{118}$

Check both:

$\sqrt{2 \left(89 - 8 \sqrt{118}\right) + 4} + \sqrt{3 \left(89 - 8 \sqrt{118}\right) - 5} = 4$

$4 = 4$

$\sqrt{2 \left(89 + 8 \sqrt{118}\right) + 4} + \sqrt{3 \left(89 + 8 \sqrt{118}\right) - 5} = 4$

$41.71 \ne 4$

We must discard the second root; the only solution is:

$x = 89 - 8 \sqrt{118}$