# 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 $x=-5-3$
$x=-2,-8$
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
$x=2,-6$

## Video-Lesson Transcript – Part 1

Let’s go over solving equations by taking square roots.

If we have this equation:

$x^2 = 9$

To solve for the value of $x$, we have to get the square root.

$\sqrt{x^2} = \sqrt{9}$

The tricky part here is the square root of $9$.

Because it’s not just positive, it can also be negative.

$x = \pm 3$

$3$ and $-3$

Let’s have a more complicated one.

$x^2 - 5 = 31$

In order to solve this, we have to leave $x^2$ by itself first.

So let’s get rid of $-5$ by adding $5$ on both sides of the equation.

$x^2 - 5 + 5 = 31 + 5$
$x^2 = 36$
$\sqrt{x^2} = \sqrt{36}$
$x = \pm 6$

So our solutions are

{$6, -6$}

It’s also possible that we get one that doesn’t work out.

$x^2 + 7 = 14$
$x^2 + 7 - 7 = 14 - 7$
$x^2 = 7$
$\sqrt{x^2} = \sqrt{7}$
$x = \pm \sqrt{7}$

So we’ll leave it at that.

$\sqrt{7}$ and $- \sqrt{7}$

Let’s have another one

$x^2 + 8 = 26$
$x^2 + 8 - 8 = 26 - 8$
$x^2 = 18$
$\sqrt{x^2} = \sqrt{18}$

We can write break this down into:

$x = \pm \sqrt{9} \sqrt{2}$

I chose to write this because square root of $9$ is possible.

$x = \pm 3 \sqrt{2}$

Our solutions are:

$3 \sqrt{2}$ and $-3 \sqrt{2}$

## Video-Lesson Transcript – Part 2

Let’s take a more in depth look at solving quadratic equations by taking square root.

Just to recap:

If we have $x^2 = 9$, to solve this

$\sqrt{x^2} = \sqrt{9}$
$x = \pm 3$

So our solution set is

{$+ 3, - 3$}

We could use the same method if we have

$(x + 5)^2 = 9$

Rather than multiplying, let’s just get the square roots of both sides.

$\sqrt{(x + 5)^2} = \sqrt{9}$
$x + 5 = \pm 3$

Now, let’s get the value of $x$

$x + 5 - 5 = \pm 3 - 5$
$x = - 5 \pm 3$

$- 5 + 3 = - 2$
and
$- 5 - 3 = - 8$

Our solution set is

{$- 2, - 8$}

Let’s look at this one

$(x + 7)^2 = 36$

Let’s do the same thing

$\sqrt{(x + 7)^2} = \sqrt{36}$
$x + 7 = \pm 6$
$x + 7 - 7 = - 7 \pm 6$

So we have

$-7 + 6 = -1$
and
$-7 - 6 = -13$

Our solution set is

{$-1, -13$}

Let’s have a different equation such as

$(x + 2)^2 - 3 = 13$

To solve this, we have to get rid of $-3$ first.

$(x + 2)^2 - 3 + 3 = 13 + 3$
$(x + 2)^2 = 16$
$\sqrt{(x + 2)^2} = \sqrt{16}$
$x + 2 = \pm 4$
$x + 2 - 2 = - 2 \pm 4$
$x = - 2 \pm 4$

$-2 + 4 = 2$
and
$-2 - 4 = -6$

Our solution set is

{$2, -6$}

Let’s see this one

$(x + 2)^2 - 3 = 11$

Let’s see what happens when we solve this

$(x + 2)^2 - 3 = 11$
$(x + 2)^2 - 3 + 3 = 11 + 3$
$(x + 2)^2 = 14$
$\sqrt{(x + 2)^2} = \sqrt{14}$

Since we can not get the square root of $14$, we’ll leave it as:

$x + 2 = \pm \sqrt{14}$

Then isolate $x$

$x + 2 - 2 = -2 \pm \sqrt{14}$
$x = -2 \pm \sqrt{14}$

{$-2 + \sqrt{14}, -2 - \sqrt{14}$}