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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

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

So we have two answers:

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.

Our final answers are

\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.

So our answer is

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

So we have two answers:

- 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

So we have two answers:

-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}

This can already be our answers.

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