# Solving Radical Equations and Inequalities

This section covers:

Now that we’ve learned about Quadratics and Factoring and did some work with square roots,  we can go back and revisit solving radical equations and inequalities, with special emphasis on square root functions.  We solved some radical equations in the Solving Exponential and Radical Equations portion in the Exponents and Radicals in Algebra section, but now we can work with more complicated equations where we can multiply binomials to find the answers!

For factoring and solving with Exponents, see the Exponential Functions section.

First of all, let’s see what some basic radical function graphs look like.   The first set of graphs are the quadratics and the square root functions; since the square root function “undoes” the quadratic function, it makes sense that it looks like a quadratic on its side.  But the important thing to note about the simplest form of the square root function  is that the range (y) by definition is only positive; thus we only see “half” of a sideways parabola.  The domain (x) is always positive, too, since we can’t take the square root of a negative number.

Remember also that another way to write  is .

See how, since  is 4, a point on the quadratic graph is (2, 4)?  Similarly, since the square root of 4 is 2, a point on the square root graph is (4, 2).

Next we have the cubic (raising something to the 3rd power) and cube root function graphs.  Since cube roots can be both positive and negative, the domain and range of both graphs is the set of real numbers.

Remember also that another way to write  is .

See how, since  is 8, a point on the cube function graph is (2, 8)?  Similarly, since the cube root of 8 is 2, a point on the cube root graph is (8, 2).   Note that (–8, –2) is also a point on this graph, since the cube root of –8  is –2.

We’ll talk a little later in the Advanced Functions: Inverses section that the quadratic and square root functions are “opposites” or inverses of each other.  The cubic and cube root functions are also inverses of each other.

Now let’s solve some problems with square root functions.  With even radicals, we have to make sure that our answers never produce a negative number underneath the square root (even radical) sign.   Also, if we raise both sides to an even exponent (like squaring), we need to check our answers, since some solutions may not work.  Both of these conditions can produce extraneous solutions (solutions that don’t work), since even exponents can be a little quirky.

The main idea in solving these is to get rid of the radical signs by raising both sides to that exponent.  For example, with square roots, we have to square both sides.  If we have two square roots, it’s easiest to have them separated so when we square both sides, it’s not as complicated.  Sometimes we have to take the square of each side more than once, after we’ve FOILED one or both sides (see last example in next set of examples).

And don’t forget when we end up with a quadratic equation, put everything on one side (set to 0) and factor, or use the quadratic formula.

Here are some of the problems we solved previously and a few more (since we know how to FOIL now!):

Here are a few where we have to square both sides two times to get rid of the radicals.  Note the second problem has a radical inside of a radical.

And are a couple of examples with odd-indexed radicals, where we can sit back and relax and just solve – everything we get should work!

We can graph radical functions either with a t-chart or in the graphing calculator.  Later, we’ll learn how to transform functions more easily in the Parent Graphs and Transformations  section.

Since we’re so good with the graphing calculator (yeah!), let’s solve a radical function equation using the calculator:

Note that we saw some of these same examples in the Exponents and Radicals in Algebra section.

Again, when solving inequalities, we need to be careful with multiplying and dividing by anything negative, where we have to change the direction of the inequality sign.

We also have to be careful that our answer still keeps what’s under an even radical to be positive (domain restriction), so we have to create another inequality and set what’s under the even radical to greater than or equal to 0, solve for x, and take the intersection of both solutions.   When we solve for x in these situations, we get the critical values.  The reason we take the intersection of the two solutions is because both must work.

There’s one more check we need to do.   Sometimes, because taking the even radical of an expression give the positive value only), we may end up with some extraneous solutions.    For example, in the inequality  $$\displaystyle \sqrt{{2-x}}<x$$  (see below for solution), since the left hand side has to be  0 (since it’s a square root),  the right hand side has to be > 0 (so x > 0).    The best way to eliminate these extraneous solutions is to check our answers in the original inequality for each interval we get to make sure they work.

For more advanced solving, we’ll want to use a sign chart to show the intervals that work and don’t work;  I have examples without using a sign chart in the first table below, and examples using them in the second table.

Here are the examples.

Here are some problems where we need to use a sign chart to solve radical inequalities.   Note:  you can check these these inequalities in your graphing calculator to make sure they are correct.  For example, for the first one use  $${{Y}_{1}}=\sqrt{{{{x}^{2}}-2x-8}}$$  and  $${{Y}_{2}}=x+2$$.

Here are a few more problems with radical inequalities:

Let’s first just graph a simple radical inequality to show what the shading looks like; you may have to make some graphs like this.  We saw earlier what the radical function looks like, and we can use the “rain up” (for >) and “rain down” (for <) shading like we did here in the Quadratics Inequalities section.  Remember that with “<” and “>” inequalities, we draw a dashed (or dotted) line to indicate that we’re not really including that line (but everything up to it), whereas with “” and “”, we draw a regular line, to indicate that we are including it in the solution.

Note that we also had to check so that anything under the even radical is positive; this is why the graph is shaded for .  We still have to keep this vertical line dotted, since we take the intersection (both have to work) of the two inequalities, and in this example, we have <.

Note that we can put this in the graphing calculator, too.  We had to move the cursor way to the left of  and change to an inequality (play around with it; it’s different with the color calculator!) and then graph:

We can also solve radical inequalities graphically.  To get the intervals for x for these graphs, you have to look and see whichever graph is on the bottom or below the other one (has the smaller y for that interval) if it is a “less than” problem.   For a “greater than” problem, you find the interval of the graph that is on the top or above the other one (has a larger y for that interval).

Sometimes (like in the third example below), there are no values that make the inequality true.

NOTE:  We could solve use graphing calculator as we did for the equalities above, but it’s really difficult to get the point of intersection where the graphs hit the x axis, since the graphs are just starting there (you might use the table).  Also, when you get the other points of intersection, it’s easiest if you have use TRACE to move the cursor above the intersection of the functions, if there is an intersection.  When you get the point of intersection, use the x value, since we’re solving for x.

Learn these rules, and practice, practice, practice!

For Practice: Use the Mathway widget below to try an Inequality problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Imaginary (Complex) Numbers – you’re ready!

## 21 thoughts on “Solving Radical Equations and Inequalities”

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the guy who loves your prefrontal cortex

2. I like your site. I wish I could’ve found it sooner it would have helped me with high school math.

• Sean – I’m sorry you didn’t find what you’re looking for – can you tell me, so maybe I can add something? Lisa

• Thanks for writing. I would divide both sides by 54, so you get 2/45=d^(-13.5/3) (I assumed the 13.5 was part of the exponent). Then raise both sides to the (-3/13.5) to get d alone. So you get (2/45)^(-3/13.5) = 1.99749, or about 2. Does that make sense? Lisa

3. When solving the inequalities with radicals on both sides and having to find where they all “intersect”, do you find the inequality statement from guess and check method? Or is there a faster way to determine the final inequality? Also, will the final inequality ever be an “or” statement, (x is greater than 2 OR x is less than 5), or will it always be an “and” statement (x is less than 6 AND greater than 3)???

• Great questions! When you are graphing, you find where they intersect, and then look which line is above the other one on either side. If it’s < for example, and the "first function" is below the "second function", then the part of the graph that you want is on the left - because that's where the statement is "true". Does that make sense? For your second questions, the final inequality will always be an "and" if you're checking the inequality AND checking to make sure what's under the radical is greater than or equal to 0. But you may have some "or" inequalities when you're dealing with quadratic inequalities and/or absolute value inequalities, and you have a >, like in some examples here.
Hope that helps! Lisa

• kindly have more example because i want to know and how to solve it by easily .. 🙂 thank you

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Lisa

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• Thanks so much for letting me know – this makes me want to hurry up and finish the site!!! Lisa

6. hey…..thanxx for the awesome content….but I do have some doubts if u could plz help me with them..it would be great

thanx.

• I get that this is only true between 1/2 and 2. I did a sign table using 1/2 and 2 (set the numerator and denominator to 0 to get critical points) and then did tests in the three areas. Does this make sense?