# Solving Equations That Have Radical Terms

## Part 1

In this video, we are going to solve simple equations containing radical expression.

In $\sqrt{x}=7$, first square both sides of the equation
$\sqrt{x}^2=7^2$
$x=49$

If there are other values in the equation, then first isolate the $\sqrt{x}$ before squaring both sides of the equation.

## Part 2

In the second part, we are going to solve more complex equations containing radical expression.

For example:
$\sqrt{x}=(x-5)$

Square both sides of the equation
$\sqrt{x}^2=(x-5)^2$

Use FOIL to complete the squaring if necessary
$x=(x-5)(x-5)$
$x=x^2-5x-5x+25$

Combine like terms
$x=x^2-10x+25$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute the variables
$x=\frac{11\pm\sqrt{(-11)^2-4(1)(25)}}{2(1)}$
$x=\frac{11\pm\sqrt{121-4(1)(25)}}{2(1)}$
$x=\frac{11\pm\sqrt{21}}{2}$

So we have:
$x=\frac{11+\sqrt{21}}{2}$ and $x=\frac{11-\sqrt{21}}{2}$

In the end, x is approximately 7.8 or 3.2

If it was -7.8 instead of 7.8, then it cannot be a negative answer since the square root of a negative number is imaginary. In this case, the final answer would just be 3.2