This section covers:
 Radical Function Graphs
 Solving Radical Equations Algebraically
 Solving Radical Equations Graphically
 Solving Radical Inequalities Algebraically
 Solving Radical Inequalities Graphically
 More Practice
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.
Radical Function Graphs
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 \(y=\sqrt{x}\) 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.
Quadratic Function  Square Root Function 
Domain: \(\left( {\infty ,\infty } \right)\text{ or }\mathbb{R}\) Range: \(\left[ {0,\text{ }\infty } \right)\) 
Domain: \(\left[ {0,\text{ }\infty } \right)\) Range: \(\left[ {0,\text{ }\infty } \right)\) 
Remember also that another way to write \(y=\sqrt{x}\) is \(y={{x}^{{\frac{1}{2}}}}\).
See how, since \({{2}^{2}}=4\), a point on the quadratic graph is \(\left( {2,4} \right)\)? Similarly, since \(\sqrt{4}=2\), a point on the square root graph is \(\left( {4,2} \right)\).
Next, we have the cubic (raising something to the 3^{rd} 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.
Cubic Function  Cube Root Function 
Domain: \(\left( {\infty ,\infty } \right)\text{ or }\mathbb{R}\) Range: \(\left( {\infty ,\infty } \right)\text{ or }\mathbb{R}\) 
Domain: \(\left( {\infty ,\infty } \right)\text{ or }\mathbb{R}\) Range: \(\left( {\infty ,\infty } \right)\text{ or }\mathbb{R}\) 
Remember also that another way to write \(y=\sqrt[3]{x}\) is \(y={{x}^{{\frac{1}{3}}}}\).
See how, since \({{2}^{3}}=8\), a point on the cube function graph is \(\left( {2,8} \right)\) (and so is \(\left( {2,8} \right)\))? Similarly, since \(\sqrt[3]{8}=2\), a point on the cube root graph is \(\left( {8,2} \right)\). Note that \(\left( {8,2} \right)\) is also a point on this graph, since \(\sqrt[3]{8}=2\).
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.
Solving Radical Equations Algebraically
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).
Also remember that you can always turn a radical into a rational (fractional) exponent; here is an example: \({{\left( {\sqrt[3]{x}} \right)}^{4}}=\sqrt[3]{{{{x}^{4}}}}={{x}^{{\frac{4}{3}}}}\).
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!):
Radical Equation  Notes 
\(\displaystyle \sqrt{x}=3\)
\(\displaystyle \begin{array}{c}{{\left( {\sqrt{x}} \right)}^{2}}={{\left( {3} \right)}^{2}}\\x=9\end{array}\)
Doesn’t work, so no solution! 
After we square each side and solve for \(x\), we get \(x=9\). Since we squared each side, we need to check our answer to see if it works.
When we take the square root of a number, we only take the positive value, so our answer doesn’t work! You can also see early on in the problem that you can’t get a negative number from a square root, so the answer is no solution, or \(\emptyset \). 
\(4\sqrt{{x1}}=\sqrt{{x+1}}\)
\(\begin{align}{{\left( {4\sqrt{{x1}}} \right)}^{2}}&={{\left( {\sqrt{{x+1}}} \right)}^{2}}\\\,{{4}^{2}}\left( {x1} \right)&=\left( {x+1} \right)\\16x16&=x+1\\15x&=17;\,\,\,\,\,x=\frac{{17}}{{15}}\end{align}\)

Since we have square roots on both sides, we can simply square both sides to get rid of them. We have to make sure we square the 4 too, since it is on the outside of the radical. Then we solve for \(x\) to get \(\displaystyle \frac{{17}}{{15}}\).
We have to make sure our answers don’t produce any negative numbers under the square root; this looks good.
Also, since we squared both sides, let’s check our answer: \(\displaystyle 4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\frac{{32}}{{15}}}}?\,\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=\sqrt{{\left( {16} \right)\left( 2 \right)\frac{1}{{15}}}}\,\,?\,\,\,\,\,\,\,\,\,\,\,\,\,4\sqrt{{\frac{2}{{15}}}}=4\sqrt{{\frac{2}{{15}}}}\,\,\,\,\surd \) 
\(\sqrt[4]{{2x}}=\sqrt[4]{{x4}}\)
\(\begin{align}{{\left( {\sqrt[4]{{2x}}} \right)}^{4}}&={{\left( {\sqrt[4]{{x4}}} \right)}^{4}}\\2x&=x4\\2x&=6\\x&=3\end{align}\)
Doesn’t work, so no solution! 
Here’s one where we have fourth roots instead of square roots. Since we have the 4^{th} root on each side, we can just raise each side to the 4^{th} power and solve.
We correctly solved the equation, but notice that when we plug in 3 in the first radical (and the second one too!), we get a negative number (\(23=1\)). We can’t take even roots of negative numbers.
We have to “throw away” our answer and the correct answer is “no solution” or \(\emptyset \). 
\(\displaystyle {{\left( {x1} \right)}^{{\frac{1}{2}}}}=x3\)
\(\require{cancel} \displaystyle \begin{array}{c}{{\left( {{{{\left( {x1} \right)}}^{{\frac{1}{2}}}}} \right)}^{2}}={{\left( {x3} \right)}^{2}}\\x1={{x}^{2}}6x+9\\{{x}^{2}}7x+10=0\\\left( {x5} \right)\left( {x2} \right)=0\\\text{x=5,}\,\,\,\cancel{{x=2}}\end{array}\)
2 doesn’t work! 
First of all, remember that \(\displaystyle {{x}^{{\frac{1}{2}}}}=\sqrt{x}\), so we still solve by squaring both sides (see that the exponents “cancel out” on the left?).
When we square each side to get rid of the radical sign, we find we have to FOIL, or multiply two binomials on the right side. Now we’re left with a quadratic equation, so we have to get everything to one side and set to 0.
When we factor, we get both 5 and 2 for answers. 5 works, but when we put 2 in the original problem, we get a negative number on the righthand side. Since the square root of something can never be a negative number, we have to eliminate this solution (so 2 is an extraneous solution). The answer is \(x=5\).

\(\displaystyle \begin{array}{c}{{\left( {\sqrt[3]{{x+2}}} \right)}^{4}}+2=18\\{{\left( {x+2} \right)}^{{\frac{4}{3}}}}=16\end{array}\)
\(\displaystyle \begin{align}{{\left( {{{{\left( {x+2} \right)}}^{{\frac{4}{3}}}}} \right)}^{{\frac{3}{4}}}}&={{16}^{{\frac{3}{4}}}}\\x+2&=\pm {{2}^{3}}\\x&=\pm {{2}^{3}}2\\x&=82=6\,\,\,\,\,\text{and}\\x&=82=10\end{align}\) 
With both a root and an exponent, turn the radical into a rational (fractional) exponent, and raise each side to the reciprocal of the exponent.
If the even number is on the top of the fraction, you have to take the positive and negative solutions. We also have to make sure the number on the right that we’re raising to an exponent is positive, or there would be no answers.
Let’s check our answers: \(\displaystyle \begin{array}{c}{{\left( {\sqrt[3]{{6+2}}} \right)}^{4}}+2={{\left( {\sqrt[3]{8}} \right)}^{4}}+2={{2}^{4}}+2=18\,\,\,\,\,\,\surd \\{{\left( {\sqrt[3]{{10+2}}} \right)}^{4}}+2={{\left( {\sqrt[3]{{8}}} \right)}^{4}}+2={{\left( {2} \right)}^{4}}+2=18\,\,\,\,\,\,\surd \end{array}\) 
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.
Radical Equation  Notes 
\(\displaystyle 5+\sqrt{{3x8}}\sqrt{{4x}}=3\)
\(\displaystyle \begin{array}{c}\sqrt{{3x8}}=\sqrt{{4x}}2\\{{\left( {\sqrt{{3x8}}} \right)}^{2}}={{\left( {\sqrt{{4x}}2} \right)}^{2}}\\\,3x8=4x4\sqrt{{4x}}+4\\\,4\sqrt{{4x}}=x+12\\{{\left( {4\sqrt{{4x}}} \right)}^{2}}={{\left( {x+12} \right)}^{2}}\\\,16\left( {4x} \right)={{x}^{2}}+24x+144\\{{x}^{2}}40x+144=0\\\left( {x4} \right)\left( {x36} \right)=0\\x=4,\,\,\,x=36\,\end{array}\)
Both work! 
Since we have two radicals, let’s put one on each side to make the squaring easier. Also, combine any like terms before squaring each side.
After squaring each side, we see that we still have a radical, so combine like terms again, and square each side once more.
Then we have a quadratic, so we need to put everything to one side and set to 0. Factor, set each factor to 0, and then get answers.
We have to make sure the answers work under the radicals (causing no negative numbers under the radicals) – this seems fine. Also, since we squared both sides, we have to check our answers in the original problem – both work. \(\displaystyle \surd \) 
\(\displaystyle \sqrt{{{{x}^{2}}2+2\sqrt{{3x+3}}}}=x+1\)
\(\displaystyle \begin{array}{c}{{x}^{2}}2+2\sqrt{{3x+3}}={{\left( {x+1} \right)}^{2}}\\\cancel{{{{x}^{2}}}}2+2\sqrt{{3x+3}}=\cancel{{{{x}^{2}}}}+2x+1\\\,2\sqrt{{3x+3}}=2x+3\\\,{{\left( {2\sqrt{{3x+3}}} \right)}^{2}}={{\left( {2x+3} \right)}^{2}}\\4\left( {3x+3} \right)=4{{x}^{2}}+12x+9\\\,\,\cancel{{12x}}+12=4{{x}^{2}}+\cancel{{12x}}+9\\\,4{{x}^{2}}=3\\\,{{x}^{2}}=\frac{3}{4}\end{array}\) \(\displaystyle x=\,\frac{{\sqrt{3}}}{2}\,\,\,\,\,\,\,\,\left( {\frac{{\sqrt{3}}}{2}\text{ doesn }\!\!’\!\!\text{ t work}} \right)\) 
Square both sides first to get rid of the outside square root. Then simplify, get the radical by itself on one side, and square both sides again.
Check both answers to make sure we have no negatives under radicals. Note that \(\displaystyle \frac{{\sqrt{3}}}{2}\) makes the outside radical negative, so it is extraneous.
The positive answer works in the original problem! \(\displaystyle \surd \)

And are a couple of examples with oddindexed radicals, where we can sit back and relax and just solve – everything we get should work!
Radical Equation  Notes 
\(2\sqrt[3]{{x+2}}=\sqrt[3]{{x+8}}\)
\(\begin{array}{c}{{\left( {2\sqrt[3]{{x+2}}} \right)}^{3}}={{\left( {\sqrt[3]{{x+8}}} \right)}^{3}}\\8\left( {x+2} \right)=x+8\\8x+16=x+8\\\,7x=8\\\,x=\frac{8}{7}\end{array}\) 
We cube both sides, not forgetting to cube the 2 since it’s on the outside of the radical sign.
We then solve for \(x\), not worrying about the sign! Much easier! 
\(\displaystyle 2\sqrt[5]{x}\sqrt[5]{{{{x}^{2}}4x8}}=0\)
\(\displaystyle \begin{array}{c}2\sqrt[5]{x}=\sqrt[5]{{{{x}^{2}}4x8}}\\{{\left( {\,2\sqrt[5]{x}} \right)}^{5}}={{\left( {\sqrt[5]{{{{x}^{2}}4x8}}} \right)}^{5}}\\32x={{x}^{2}}4x8\\{{x}^{2}}36x8=0\end{array}\) \(\displaystyle x=\frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}=\frac{{36\pm \sqrt{{1328}}}}{2}\) \(\displaystyle =\frac{{36\pm 4\sqrt{{83}}}}{2}=18\pm 2\sqrt{{83}}\) 
First, move one of the radicals to the other side so we can raise each side to the 5^{th} power. We end up with a quadratic to solve.
Sorry – I snuck in one where we couldn’t factor – so we have to use the quadratic formula to solve the quadratic.
You can actually check this answer (or any of these) by storing the solution(s) in \(x\) in your calculator.
For example, to check in the positive value, type \(18+2\sqrt{{83}}\), and then hit STO> X,T,,t. Then type in \(2x\hat{\ }(1/5)(x\hat{\ }24x8)\hat{\ }(1/5)\), using thebutton for \(x\). You should get 0. 
Solving Radical Equations Graphically
We can graph radical functions either with a tchart 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:
Radical Equation and Graph  Calculator Screens 
\(\sqrt{{5x16}}=\sqrt{{2x4}}\)


Notes: Push Y = and enter the two equations in Y_{1} = and Y_{2} =, respectively.
Push GRAPH. You may need to hit “ZOOM 6” (ZoomStandard) and/or “ZOOM 0” (ZoomFit), and maybe “ZOOM 3” (Zoom Out) ENTER to make sure you see the lines crossing in the graph.
Hit TRACE and then use the round right arrow to move the cursor up to one of the graphs above the intersection (otherwise, the Intersect may not work with these square root functions). You may have to do this again for the other curve after the first ENTER (Second Curve?) below.
To get the point of intersection, push “2^{nd} TRACE” (CALC), and then push 5 (for intersect) (First Curve?), ENTER (Second Curve?), ENTER (Guess?), ENTER. You should see the point of intersection on the bottom!
Now we are solving for the \(x\) in these types of problems, so our answer is 4. 
Solving Radical Inequalities Algebraically
Note that we saw some of these same examples in the Exponents and Radicals in Algebra section.
Remember these rules for solving inequalities algebraically:
 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.
 What’s under an even radical has to be positive (domain restriction); we have to create another inequality and set what’s under the even radical to \(\boldsymbol{{\ge 0}}\). We then solve for \(x\), and take the intersection of both solutions. The reason we take the intersection of the two solutions is because both must work.
 For more advanced solving, we’ll want to use a sign chart to show the intervals that work and don’t work; when we solve for \(x\) in these situations, we get the critical values for the sign chart. We then have to check each interval to see if the inequality is true or false.
 For radicals, when there is a variable not under a square root and we square both sides, we have to be careful since don’t know if the side without the square is positive or negative (and thus if we should switch the sign). In these cases, check the interval test values in the original inequality, and use T or F (or Y or N) to indicate whether or not they work.
 If we get something like \(\sqrt{n}<0\) (or a negative number), there is no solution, and something like \(\sqrt{n}\ge 0\) (or a negative number), we get all real numbers, except for the domain restriction (\(n\ge 0\)).
 You can check these inequalities in your graphing calculator to make sure they are correct. For example, for the first one use \({{Y}_{1}}=\sqrt{{{{x}^{2}}2x8}}\) and \({{Y}_{2}}=x+2\).
Here are some examples; note that we just raise each side to the root to get rid of it. We can do that in these examples, since we know the sign of the values on both sides. It gets trickier when we don’t know the sign of one of the sides.
Radical Inequality  Notes 
\(\sqrt{{x+2}}\le 4\)
\(\begin{array}{c}\sqrt{{x+2}}\le 4\text{ }\,\text{ }\,\text{and }x+2\ge 0\\{{\left( {\sqrt{{x+2}}} \right)}^{2}}\le {{4}^{2}}\text{ }\,\,\text{and }x+2\ge 0\text{ }\\x+2\le 16\text{ }\,\text{and }x\ge 02\text{ }\\x\le 14\text{ }\,\text{and }x\ge 2\\\\\{x:2\le x\le 14\}\text{ or }\left[ {2,14} \right]\end{array}\) 
We have to solve two inequalities, since our \(x\) must work in the original, but also work so anything under the even radical is positive (domain restriction): \(\displaystyle x+2\ge 0\).
To get rid of the square roots, we square each side, and we can leave the inequality signs the same since we’re multiplying by positive numbers. Then we just solve for \(x\), just like we would for an equation.
We need to check our answer by trying random numbers in our solution (like \(x=2\)) in the original inequality (which works). We also need to try numbers outside our solution (like \(x=6\) and \(x=20\)) and see that they don’t work. 
\(\displaystyle \sqrt{{5x16}}<\sqrt{{2x4}}\)
\(\displaystyle \begin{array}{c}{{\left( {\sqrt{{5x16}}} \right)}^{2}}<{{\left( {\sqrt{{2x4}}} \right)}^{2}}\\5x16<2x4\\3x<12\\x<4\end{array}\)
Also: \(\displaystyle \begin{array}{c}5x16 \,\ge 0\text{ and 2}x4 \,\ge 0\\x\ge \frac{{16}}{5}\text{ and }x\ge 2\\x<4\,\,\,\cap \,\,\,x\ge \frac{{16}}{5}\,\,\,\cap \,\,\,x\ge 2\end{array}\) \(\displaystyle \{x:\,\,\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},\,\,4} \right)\) 
Now we have to solve three inequalities: one for the main problem, and one each for the even root radicands, that have to be \(\ge 0\). Since both sides are positive (even roots), we can safely take the square of each side.
We need to take the intersection (all must work) of the inequalities: \(\displaystyle x<4\) and \(\displaystyle x\ge \frac{{16}}{5}\) and \(\displaystyle x\ge 2\). This will give us \(\displaystyle \frac{{16}}{5}\le \,\,x<4\). (Try it yourself on a number line). Watch out for the hard and soft brackets.
We need to check our answer by trying \(x=3.5\) in the original inequality (which works) and \(x=3\) or \(x=5\) (which don’t work). 
\(\sqrt{{x+6}}\le 2\)
\(\{\}\text{ or }\emptyset \) 
Before we even need to get started with this inequality, we can notice that the square root of anything can never \(\displaystyle \le 2\) (or \(\boldsymbol{<0}\)), by definition.
We know right away that the answer is no solution, or \(\{\}\text{ or }\emptyset \). 
\(\sqrt[3]{{x3}}>4\)
\(\begin{array}{c}{{\left( {\sqrt[3]{{x3}}} \right)}^{3}}>{{4}^{3}}\,\,\,\,\\x3>64\\x>67\end{array}\) 
With odd roots, we don’t have to worry about checking underneath the radical sign, since we could have positive or negative numbers as a radicand. 
Here are more complicated problems where we need to use a sign chart to solve radical inequalities.
Note that we have to be careful when there is a variable on a side and it’s not under a square root; when we square both sides, we’re not really sure if we’d have to switch the inequality sign. This is because we don’t know if the side without the square is positive or negative. Since this is the case, it’s best to check the intervals in the original inequality, and use T or F (or Y or N) when checking.
Radical Inequality  Notes 
\(\sqrt{{{{x}^{2}}2x8}}>x+2\)
\(\displaystyle \begin{array}{c}\sqrt{{{{x}^{2}}2x8}}\,\,>x+2\,\,\text{and }{{x}^{2}}2x8\ge 0\\{{\left( {\sqrt{{{{x}^{2}}2x8}}} \right)}^{2}}>{{\left( {x+2} \right)}^{2}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{and}\,\,\left( {x+2} \right)\left( {x4} \right)\ge 0\text{ }\\\cancel{{{{x}^{2}}}}2x8>\cancel{{{{x}^{2}}}}+4x+4\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{and }\left( {x+2} \right)\left( {x4} \right)\ge 0\text{ }\\\,6x>12\text{ and }\left( {x+2} \right)\left( {x4} \right)\ge 0\\\,\,\,\,x<2\text{ }\,\text{and }\left( {x+2} \right)\left( {x4} \right)\ge 0\end{array}\)
Here is the sign chart for the righthand inequality, including the restriction for the lefthand:

We have to solve two inequalities, since our \(x\) must work in the original, but also work so anything under the even radical is positive (domain restriction): \(\displaystyle {{x}^{2}}2x8\ge 0\).
When we solve, we’ll get the critical values, or critical points. In this case, we have –2, –2 again, and 4 (use closed circles for \(\ge \), \(\le \) and open circles for just \(>,<\)).
Since \(x\) has to be less than –2, we can exclude the other intervals.
But, using the sign chart, it’s a good idea to check each interval with random points to see if the original equation works (T or F).
The answer is \(\left( {\infty ,2} \right)\). 
\(\displaystyle \sqrt{{2x}}<x\)
\(\begin{array}{c}\sqrt{{2x}}<x\text{ and }\,\,2x\ge 0\\{{\left( {\sqrt{{2x}}} \right)}^{2}}<{{x}^{2}}\,\,\,\text{and }x\le 2\text{ }\\\,2x<{{x}^{2}}\text{ }\,\text{and }x\le 2\text{ }\\{{x}^{2}}+x2<0\text{ }\,\text{and }x\le 2\text{ }\\\left( {x+2} \right)\left( {x1} \right)<0\text{ }\,\text{and }x\le 2\text{ }\end{array}\)
Here is the sign chart for the lefthand side, including the restriction for the righthand:

Again, we need two inequalities: one for the original inequality, and one for the expression under the square (even) root (domain restriction).
When we solve, we’ll get the critical values, or critical points. In this case, we have –2, 1, and 2 (use closed circles for \(\ge \), \(\le \) and open circles for just \(>,<\)).
At this point, we need to check the original equation, since we don’t know if the righthand side of the original inequality is positive or negative, which would create a sign change when solving.
So, for the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign and the square root will always be positive.
The answer is \(\left( {1,2} \right]\). Try it in the graphing calculator – it works! √ 
\(3\sqrt{{{{x}^{2}}9}}+2<14\)
\(\begin{array}{c}3\sqrt{{{{x}^{2}}9}}+2<14\,\,\,\text{and}\,\,\,{{x}^{2}}9\ge 0\\\sqrt{{{{x}^{2}}9}}<4\,\,\,\,\,\text{and}\,\,\,\left( {x3} \right)\left( {x+3} \right)\ge 0\\{{\left( {\sqrt{{{{x}^{2}}9}}} \right)}^{2}}<{{4}^{2}}\,\,\,\text{and}\,\,\,\left( {x3} \right)\left( {x+3} \right)\ge 0\\{{x}^{2}}<25\,\,\,\,\,\text{and}\,\,\,\left( {x3} \right)\left( {x+3} \right)\ge 0\\\left( {x5} \right)\left( {x+5} \right)<0\,\,\,\,\,\text{and}\,\,\,\left( {x3} \right)\left( {x+3} \right)\ge 0\end{array}\)

We have to solve two inequalities, since our \(x\) must work in the original, but also work so anything under the even radical is positive (domain restriction): \(\displaystyle {{x}^{2}}9\ge 0\), or \(\left( {x3} \right)\left( {x+3} \right)\ge 0\).
We also see that \({{x}^{2}}<25\), or \(\left( {x5} \right)\left( {x+5} \right)<0\).
We’ll use a sign chart with all four critical values and check the intervals by checking original (simplified) inequality: \(\sqrt{{{{x}^{2}}9}}<4\). Note the open and close circles, used for inclusion (\(<\))or exclusion (\(\ge \)).
The answer is \(\left( {5,3} \right]\cup \left[ {3,5} \right)\). 
Here are a few more problems with radical inequalities:
Radical Inequality  Notes 
\(\sqrt[4]{x}>\sqrt[3]{x}\)
\(\displaystyle \begin{array}{c}{{x}^{{\frac{1}{4}}}}>{{x}^{{\frac{1}{3}}}}\text{ }\,\text{and}\text{ }\,\,x\ge 0\\{{\left( {{{x}^{{\frac{1}{4}}}}} \right)}^{{12}}}>{{\left( {{{x}^{{\frac{1}{3}}}}} \right)}^{{12}}}\text{ and}\text{ }\,x\ge 0\\{{x}^{3}}>{{x}^{4}}\text{ }\,\text{and}\,\text{ }\,x\ge 0\\{{x}^{4}}{{x}^{3}}<0\text{ }\,\text{and}\,\text{ }\,x\ge 0\\{{x}^{3}}\left( {x1} \right)<0\text{ }\,\text{and }x\ge 0\text{ }\end{array}\)

For this inequality, let’s first turn the roots into exponents so we can more easily get rid of them. We also have to remember to set \(x\ge 0\) since \(x\) is under an even radical (4^{th} root).
Raise each side to the 12^{th} power, since that will cancel out both of the fractional exponents (neat trick – like a common denominator for the exponents!) Note that we can safely do this, since \(x\) has to be positive (because of even root), so both sides will be positive.
When we solve, we’ll get the critical values, or critical points. In this case, we have 0 and 1. We need an open circle on 0 because of the primary inequality.
For the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign.
The answer is \(\left( {0,\,\,1} \right)\). 
\(\sqrt{x}\ge 3\sqrt[4]{x}\,\)
\(\displaystyle \begin{array}{c}\sqrt{x}\,\,\le \,\,3\,\sqrt[4]{x}\,\text{ and }\,\,x\ge 0\\{{\left( {\sqrt{x}} \right)}^{4}}\ge \,\,{{\left( {3\,\sqrt[4]{x}} \right)}^{4}}\,\text{ and }\,x\ge 0\\{{x}^{2}}\ge 81x\text{ }\,\,\text{and }\,\,x\ge 0\\{{x}^{2}}81x\ge 0\,\,\text{ and }\,\,x\ge 0\\x\left( {x81} \right)\ge 0\,\text{ and }\,x\ge 0\end{array}\)

Raise each side to the 4^{th} power, since that will cancel out both of the fractional exponents (neat trick – like a common denominator for the exponents!) Note that we can safely do this since both sides (even roots) have to be positive.
We see that the critical values are 0 and 81. Use close circles because of the \(\ge \).
For the sign chart, we check each interval with random points to see if the original equation works (T or F). Don’t forget that we can’t have a negative number under the even radical sign.
The answer is \(\left[ {81,\,\,\infty } \right)\).

Solving Radical Inequalities Graphically
Graphing 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 “\(\le \)” and “\(\ge \)”, 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 \(x\ge 5\). 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 “\({{\text{Y}}_{1}}=\)” and change to an inequality (play around with it; it’s different with the color calculator!) and then graph:
Radical Inequality and Graph  Calculator Screens 
\(y<\sqrt{{x5}}\)


Solving Radical Inequalities Graphically
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.
Radical Inequalities and Graphs  
\(\sqrt{{x+2}}\le 4\)
\(\{x:2\le x\le 14\}\text{ or }\left[ {2,14} \right]\) 
\(\sqrt{{5x16}}\,\,<\,\,\sqrt{{2x4}}\)
\(\displaystyle \,\{x:\frac{{16}}{5}\le x<4\}\text{ or }\left[ {\frac{{16}}{5},4} \right)\) 
\(\,\sqrt{{x+6}}\le 2\)
\(\,\{\}\text{ or }\emptyset \) 
\(\sqrt[3]{{x3}}>4\)
\(\displaystyle \{x:x>67\}\text{ or }\left( {67,\,\,\,\infty } \right)\) 
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 Solving Absolute Value Equations and Inequalities – you’re ready!