# How do you solve -4 sqrt(x+9) = 20 and find any extraneous solutions?

Aug 3, 2016

$x \in \emptyset$

#### Explanation:

Your radical equation has no solutions for real numbers. Here's why.

Isolate the square root on one side of the equation by dividing both sides by $- 4$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 4}}} \cdot \sqrt{x + 9}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 4}}}} = \frac{20}{- 4}$

$\sqrt{x + 9} = - 5$

By definition, the square root of a positive real number must always produce a positive value. This implies that you must have

• $x + 9 \ge 0 \to$ you must take the square root of a positive number when working in $\mathbb{R}$

• $\textcolor{red}{\cancel{\textcolor{b l a c k}{- 5 \ge 0}}} \to$ the square root of a positive number must be a positive number

In your case ,$- 5$ is clearly not positive, which implies that your equation has no solutions when working in $\mathbb{R}$. Simply put, you can't find a value of $x \in \mathbb{R}$ that can satisfy both conditions.

You can have $x + 9 \ge 0$ for $x \in \left[- 9 , + \infty\right)$, but you will never have $- 5 \ge 0$, regardless of the value of $x \in \left[- 9 , + \infty\right)$.

You can write this as

$x \in \emptyset \to$ no solutions in $\mathbb{R}$