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

Apr 12, 2015

First of all, let's write the existence conditions for the two square roots, that are:

$3 x + 3 \ge 0 \Rightarrow x \ge - 1$

$x + 2 \ge 0 \Rightarrow x \ge - 2$.

The solution of that system is: $x \ge - 1$.

Before we square both the members of the equation, we have to be sure that both members are positive or zero. They are! The first member is a sum of two square roots, that, where they exist, they are positive or zero. The second member is a positive number... so:

${\left(\sqrt{3 x + 3} + \sqrt{x + 2}\right)}^{2} = {5}^{2} \Rightarrow$

$3 x + 3 + 2 \sqrt{3 x + 3} \sqrt{x + 2} + x + 2 = 25 \Rightarrow$

$2 \sqrt{\left(3 x + 3\right) \left(x + 2\right)} = 20 - 4 x \Rightarrow$

$\sqrt{\left(3 x + 3\right) \left(x + 2\right)} = 10 - 2 x$.

Now we have to square again and we have to be sure that the two members are positive... again. The first member is positive or zero because it is the product of two polynomials that before they were! The second member is positive:

$10 - 2 x \ge 0 \Rightarrow x \le 5$, that, with the previous condition $x \ge - 1$, gives a new condition:

$- 1 \le x \le 5$.

Now:

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

$3 {x}^{2} + 6 x + 3 x + 6 = 100 - 40 x + 4 {x}^{2} \Rightarrow$

${x}^{2} - 49 x + 94 = 0$

and so:

$\Delta = {b}^{2} - 4 a c = {49}^{2} - 4 \cdot 1 \cdot 94 = 2025 = {45}^{2}$

and:

${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{49 \pm 45}{2}$

and so:

${x}_{1} = \frac{49 - 45}{2} = \frac{4}{2} = 2$

${x}_{2} = \frac{49 + 45}{2} = \frac{94}{2} = 47$

BUT

$x = 2$ is the only acceptable solution, because it is in $- 1 \le x \le 5$ and $x = 47$ doesn't.