# How do you solve  (sqrt(x + 7)) - 2(sqrt(x)) = -2 and find any extraneous solutions?

Aug 4, 2016

$x = 9$

#### Explanation:

The first thing to do here is write out the conditions that a value of $x$ must satisfy in order to be a valid solution to the equation

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

You're working with real numbers, so right from the start you know that

• you can only take the square root of a positive number
• the square root of a positive number will always return a positive number

In your case, this means that you must have

$x + 7 \ge 0 \implies x \ge - 7$

$x \ge 0$

Combine these two conditions to get

$x \in \left[- 7 , + \infty\right) \cap \left[0 , + \infty\right) \implies \textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{x \in \left[0 , + \infty\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, your goal here is to get rid of the radical terms. To do that, square both sides of the equation

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

This will get you

$x + 7 - 4 \cdot \sqrt{\left(x + 7\right) \cdot x} + 4 x = 4$

Isolate the remaining radical term on one side of the equation

$- 4 \cdot \sqrt{\left(x + 7\right) \cdot x} = - 5 x - 3$

$4 \cdot \sqrt{\left(x + 7\right) + x} = 5 x + 3$

Square both sides of the equation again

${\left(4 \cdot \sqrt{\left(x + 7\right) \cdot x}\right)}^{2} = {\left(5 x + 3\right)}^{2}$

$16 \cdot \left({x}^{2} + 7 x\right) = 25 {x}^{2} + 30 x + 9$

$9 {x}^{2} - 82 x + 9 = 0$

Use the quadratic formula to find the two roots

${x}_{1 , 2} = \frac{- \left(82\right) \pm \sqrt{{\left(- 82\right)}^{2} - 4 \cdot 9 \cdot 9}}{2 \cdot 9}$

${x}_{1 , 2} = \frac{82 \pm \sqrt{6400}}{18}$

${x}_{1 , 2} = \frac{82 \pm 80}{18} \implies \left\{\begin{matrix}{x}_{1} = \frac{82 + 80}{18} = 9 \\ {x}_{2} = \frac{82 - 80}{18} = \frac{1}{9}\end{matrix}\right.$

Since you have

${x}_{1} , {x}_{2} \in \left[0 , + \infty\right)$

both roots can be valid solutions to the original equation. Do a quick calculation to see if both roots are indeed valid solutions

$\sqrt{9 + 7} - 2 \cdot \sqrt{9} = - 2$

$4 - 2 \cdot 3 = - 2 \text{ } \textcolor{g r e e n}{\sqrt{}}$

$\sqrt{\frac{1}{9} + 7} - 2 \cdot \sqrt{\frac{1}{9}} = - 2$

$\sqrt{\frac{64}{9}} - 2 \cdot \frac{1}{3} = - 2$

$\frac{8}{3} - 2 \cdot \frac{1}{3} = - 2 \text{ } \textcolor{red}{\times}$

As you can see, $x = \frac{1}{9}$ is not a solution to the original equation, i.e. it's an extraneous solution. Therefore, you can say that the original equation has one valid solution, $x = 9$.