# Solving Inequalities Involving Square Roots

Hi, and welcome to this video on Solving Inequalities with Square Roots.

When solving inequalities with square roots, we need to follow two simple steps. The first step is to solve the inequality for x the same way we would any other inequality. There’s nothing special about this step. But the second step is unique to inequalities with even roots. Today, we’ll be looking at square roots specifically. The second step is to set whatever is under the radical symbol greater than or equal to zero and solve for x again. We do this because you cannot have a negative value of x under the square root. In other words, you can’t take the square root of a negative number. This will then give us two inequalities that we then combine to make a compound inequality if the signs are opposite. If the signs are the same, we use the more restrictive inequality.

Let’s look at a few examples to see what I’m talking about.

$$\sqrt{x+3}$$>$$4$$

First, we want to solve this inequality for x normally. We do this by following the same steps we would as if we were solving an equation instead of an inequality.

We’re going to start by squaring both sides.

$$x+3$$>$$16$$

And then we’re gonna subtract 3 from both sides.

$$x$$>$$13$$

This gives us x is greater than 13.

Now it’s time for our special step. We want to set what is under the radical symbol greater than or equal to 0 and solve for x again.

So x plus 3 is under the radical symbol.

So we’ll set x plus 3 is greater than or equal to 0

$$x+3$$≥$$0$$

So now we’re going to subtract 3 from both sides.

$$x$$≥$$-3$$

Which gives us x is greater than or equal to negative 3.

Since our signs are in the same direction for both inequalities, we choose the more restrictive one as our final answer. $$x$$>$$13$$ is more restrictive than $$x$$≥$$-3$$ because, according to the second inequality, 0 could be included because it is greater than or equal to -3, but it is not truly in our solution set because it is not greater than 13. Therefore, we only want to consider values of x that are greater than 13.

Let’s try another one.

Solve for x.

$$\sqrt{2x-5}+7$$<$$12$$

We’re gonna start by subtracting 7 from both sides.

$$\sqrt{2x-5}$$<$$5$$

Then we’re gonna square both sides.

$$2x-5$$<$$25$$

$$2x$$<$$30$$

And divide by 2 on both sides.

$$x$$<$$15$$

Next, we’re gonna set $$2x–5$$ greater than or equal to 0, and solve for x again.

$$2x-5$$≥$$0$$

So we’re gonna add 5 to both sides.

$$2x$$≥$$5$$

And divide by 2 on both sides.

$$x$$≥$$\frac{5}{2}$$

In this example, our inequality signs face opposite directions, so we want to combine them to make a compound inequality. Our final answer is: $$\frac{5}{2}$$≤$$x$$<$$15$$.

Let’s try one more example together.

$$\sqrt{-3x+1}-9$$≥$$11$$

Solve for x normally.

$$\sqrt{-3x+1}$$≥$$20$$

$$-3x+1$$≥$$400$$

$$-3x$$≥$$399$$

$$x$$≤$$-133$$

Then solve so what is under the radical is not negative.

$$-3x+1$$≥$$0$$

$$-3x$$≥$$-1$$

$$x$$≤$$\frac{1}{3}$$

Since the signs are in the same direction, we choose the more restrictive case as our final answer: $$x$$≤$$-133$$.

I want to do one more problem before we go, but this time try it on your own. Once I give you the problem, pause the video and solve the inequality yourself. Then, press play and see if your answer matches up with mine.

$$\sqrt{4x+4}+11$$<$$21$$

Think you’ve got it? Let’s see!

First, we want to solve our inequality normally.

$$\sqrt{4x+4}$$<$$10$$

$$4x+4$$<$$100$$

$$4x$$<$$96$$

$$x$$<$$24$$

Then we set what is under the radical greater than or equal to 0 and solve for x again.

$$4x+4$$≥$$0$$

$$4x$$≥$$-4$$

$$x$$≥$$-1$$

Since our signs are in opposite directions, our final inequality will be a compound inequality: $$-1$$≤$$x$$<$$24$$.

Remember, we cannot take the square root of a negative value, which is why we place the expression under the radical to be greater than or equal to 0. This is true of any even root, like a fourth root or an eighth root. However, it is okay to take an odd root of a negative value. For instance, the cube root of -8 is -2 because $$-2 \times -2 \times -2 = -8$$.

If you are asked to solve an inequality with a different even root, follow the same steps we did here. If it is an odd root, you do not have to worry about the second step and can stop after you initially solve the inequality.

I hope this video on solving inequalities with square roots was helpful. Thanks for watching and happy studying!

## Solving Inequalities with Square Roots Practice Questions

Question #1:

Solve this inequality for x.
$$\sqrt{x-5}$$ < $$1$$

$$x$$ < $$6$$

$$x≥5$$

$$-5≤x$$ < $$6$$

$$5≤x$$ < $$6$$

When solving an inequality, we follow the same steps as if we are solving an equation. Because the value under the radical cannot equal zero, we must add an additional step by setting what is under the radical greater than or equal to zero and solve for x.

$$\sqrt{x-5}$$ < $$1$$
$$(\sqrt{x-5})^2$$ < $$1^2$$
$$x-5$$ < $$1$$
$$x$$ < $$6$$

$$x-5≥0$$
$$x≥5$$

Since the signs are opposite directions, our answer will be a compound inequality $$5≤x$$ < $$6$$.

Question #2:

Solve this inequality for x.
$$\sqrt{3x+9}>12$$

$$x≥-3$$
$$x>45$$

$$-3≤x$$ < $$45$$

$$3≤x$$ < $$45$$

When solving an inequality, we use the same steps as when solving an equation. Since we cannot have a zero under the radical, we have an extra step to ensure our solution set does not include any value for x that would make the value under the radical zero.

$$\sqrt{3x+9}>12$$
$$(\sqrt{3x+9})^2>12^2$$
$$3x+9>144$$
$$3x>135$$
$$x>45$$

$$3x+9≥0$$
$$3x≥-9$$
$$x≥-3$$

Since the signs are in the same direction, we choose the answer that is more restrictive, which is $$x>45$$.

Question #3:

Solve this inequality for x.
$$\sqrt{5x-1}-2$$ < $$11$$

$$x≥\frac{1}{5}$$

$$x$$ < $$34$$

$$-\frac{1}{5}≤x$$ < $$34$$

$$\frac{1}{5}≤x$$ < $$34$$

We will solve the inequality following the same steps as when solving an equation. Since we cannot have a zero under the radical, we must follow an additional step to ensure our solution set does not include any value that would make the value under the radical zero.

$$\sqrt{5x-1}-2$$ < $$11$$
$$\sqrt{5x-1}$$ < $$13$$
$$(\sqrt{5x-1})^2$$ < $$13^2$$
$$5x-1$$ < $$169$$
$$5x$$ < $$170$$
$$x$$ < $$34$$

$$5x-1≥0$$
$$5x≥1$$
$$x≥\frac{1}{5}$$

Since the signs are in opposite directions, our answer will be in the form of a compound inequality, which is $$\frac{1}{5}≤x$$ < $$34$$.

Question #4:

Solve this inequality for x.
$$\sqrt{6x+10}≥8$$

$$x≥-\frac{5}{3}$$
$$x≥9$$
$$-\frac{5}{3}≤x≤9$$
$$\frac{5}{3}≤x≤9$$

When solving an inequality, we follow the same steps as if we are solving an equation. However, since the value under the radical cannot be zero, we follow an additional step to ensure the solution set does not include any value that would make the value under the radical zero.

$$(\sqrt{6x+10})≥8$$
$$(\sqrt{6x+10})^2≥8^2$$
$$6x+10≥64$$
$$6x≥54$$
$$x≥9$$

$$6x+10≥0$$
$$6x≥-10$$
$$x≥-\frac{5}{3}$$

Since the signs are in the same direction, we choose the answer that is more restrictive and does not include zero, which is $$x≥9$$.

Question #5:

Solve this inequality for x.
$$\sqrt{7x+4}-5≥7$$

$$x≥-\frac{4}{7}$$
$$x≥20$$
$$-\frac{4}{7}≤x≤20$$
$$\frac{4}{7}≤x≤20$$

We will solve the inequality the same way as we would solve an equation. Since this inequality has a radical, we must follow an additional step to ensure our solution set does not include zero.

$$\sqrt{7x+4}-5≥7$$
$$\sqrt{7x+4}≥12$$
$$(\sqrt{7x+4})^2≥12^2$$
$$7x+4≥144$$
$$7x≥140$$
$$x≥20$$

$$7x+4≥0$$
$$7x≥-4$$
$$x≥-\frac{4}{7}$$

Since the signs are in the same direction, we choose the answer that is more restrictive, which is $$x≥20$$.