# How do you solve sqrt(3x+4)-sqrt(2x-7)=3?

May 30, 2017

$x \in \left\{4 , 64\right\}$

#### Explanation:

We have:

$\sqrt{3 x + 4} = 3 + \sqrt{2 x - 7}$

Squaring both sides, we get:

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

$3 x + 4 = 9 + 6 \sqrt{2 x - 7} + 2 x - 7$

Regrouping non-square root terms to one side of the equation, we get:

$x + 2 = 6 \sqrt{2 x - 7}$

Square again:

${x}^{2} + 4 x + 4 = 36 \left(2 x - 7\right)$

${x}^{2} + 4 x + 4 = 72 x - 252$

${x}^{2} - 68 x + 256 = 0$

$\left(x - 4\right) \left(x - 64\right) = 0$

$x = 4 \mathmr{and} 64$

Now test to see whether or not the solutions are valid.

Testing $x = 4$

sqrt(3(4) + 4) =^? 3 + sqrt(2(4) - 7)

4 = 3 + 1 color(green)(√)

Testing $x = 64$

sqrt(3(64) + 4) =^? 3 + sqrt(2(64) - 7)

sqrt(196) =^? 3 + sqrt(121)

14 = 3 + 11 color(green)(√)

So our solution set is $x \in \left\{4 , 64\right\}$.

Hopefully this helps!