# How do you solve sqrt(x+4) +sqrt(2x-1)=3sqrt(x-1)?

Jun 6, 2016

#### Explanation:

Before solving, let's determine the restrictions on the variable. A radical is undefined it the number inside is less than 0.

We must therefore set up an inequality and solve:

$\sqrt{x + 4} \ge 0$

$\sqrt{2 x - 1} \ge 0$

and

$3 \sqrt{x - 1} \ge 0$

Solving each we get that

$x \ge - 4$

$x \ge \frac{1}{2}$

and

$x \ge 1$

We must pick the largest of these, since if we pick one of the smaller ones the equation will become undefined. The largest is $x \ge 1$; this is our restriction statement. To justify the point I made earlier in this paragraph, try substituting a number smaller than $1$ into the equation. For example, $x = - 2$

$\sqrt{- 2 + 4} + \sqrt{2 \times - 2 - 1} = 3 \sqrt{- 2 - 1}$

$\sqrt{2} + \sqrt{- 5} = 3 \sqrt{- 3}$

In this case, two radicals are undefined. Now that we have defined any restrictions on the variable, we can proceed with solving. This can be started by squaring both sides

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

$x + 4 + 2 \sqrt{x + 4} \sqrt{2 x - 1} + 2 x - 1 = 9 \left(x - 1\right)$

$3 x + 3 + 2 \sqrt{x + 4} \sqrt{2 x - 1} = 9 x - 9$

$2 \sqrt{x + 4} \sqrt{2 x - 1} = 6 x - 12$

$\sqrt{x + 4} \left(\sqrt{2 x - 1}\right) = \frac{\cancel{2} \left(3 x - 6\right)}{\cancel{2}}$

Since the radicals are both square roots, we can combine them in multiplication.

$\sqrt{\left(x + 4\right) \left(2 x - 1\right)} = 3 x - 6$

$\sqrt{2 {x}^{2} + 8 x - x - 4} = 3 x - 6$

${\left(\sqrt{2 {x}^{2} + 7 x - 4}\right)}^{2} = {\left(3 x - 6\right)}^{2}$

$2 {x}^{2} + 7 x - 4 = 9 {x}^{2} - 36 x + 36$

$0 = 7 {x}^{2} - 43 x + 40$

$0 = 7 {x}^{2} - 8 x - 35 x + 40$

$0 = 7 x \left(x - \frac{8}{7}\right) - 35 \left(x - \frac{8}{7}\right)$

$0 = \left(7 x - 35\right) \left(x - \frac{8}{7}\right)$

$x = 5 \mathmr{and} \frac{8}{7}$

Checking both answers in the original equation, we find that only $x = 5$ works.

Therefore, our solution set is $\left\{x = 5 , x \ge 1\right\}$

Hopefully this helps!