# Solving an Equation by Taking the Square Root

## Part 1

In this video, we are going to look at solving equations by taking the square root.
For example:
If we are given the equation
$x^2=9$
we can solve for x by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{9}$
This will leave us with just x. However, the square root of 9 isn’t just 3. It can be positive or negative 3. So
$x=\pm{3}$
If we had something more complicated like
$x^2-5=31$
then we would first have to get the $x^2$ by itself. So, first add 5 to both sides to isolate the $x^2$. Now we are left with $x^2=36$. Then take the square root of both sides
$\sqrt{x^2}=\sqrt{36}$
$x=\pm6$

## Part 2

In this video, we are going to look at solving equations by taking the square root more in depth.
For example:
If we are given the equation
$(x+5)^2=9$
we can first take the square root of both sides.
$\sqrt{(x+5)^2}=\sqrt{9}$
This will leave us with
$x+5=\pm{3}$
Then subtract 5 from both sides and get
$x=-5+3$ and $x=-5-3$
This leads us to a final answer of
$x=-2,-8$
If we had something more complicated like
$(x+2)^2-3=13$
then we would first have to get the $(x+2)^2$ by itself. So, first add 3 to both sides to isolate the $(x+2)^2$. Now we are left with $(x+2)^2=16$. Then take the square root of both sides
$\sqrt{(x+2)^2}=\sqrt{16}$
$x+2=\pm4$
Subtract 2 from both sides
$x=-2\pm4$
$x=-2+4$ and $x=-2-4$
So
$x=2,-6$

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