How to solve equations with radicals
Our plan for solving radical equations
In this lesson we’ll look at how to solve for the variable in a radical equation by isolating the radical, squaring both sides and then using inverse operations.
The thing to remember about solving a radical equation is that if you can get the radical term by itself, then you just need to square both sides and solve for the variable.
Let’s look at a few examples.
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Example
Solve for the variable.
???\sqrt{x}-3=2???
We have to keep the equation balanced, so when we add ???3??? to the left side, we’ll also add it to the right side.
???\sqrt{x}-3=2???
???\sqrt{x}-3+3=2+3???
???\sqrt{x}=5???
Squaring both sides, we get
???\left(\sqrt{x}\right)^2=5^2???
???x=25???
Examples of solving equations with radicals (roots)
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Always try to isolate the root by itself on one side of the equation, if possible
Example
Solve for the variable.
???\sqrt{x-2}+5=9???
We have to keep the equation balanced, so when we subtract ???5??? from the left side, we’ll also subtract it from the right side.
???\sqrt{x-2}+5=9???
???\sqrt{x-2}+5-5=9-5???
???\sqrt{x-2}=4???
Squaring both sides, we get
???\left(\sqrt{x-2}\right)^2=4^2???
???x-2=16???
Now add ???2??? to both sides.
???x=16+2???
???x=18???
Let’s look at another example with an ???x^2??? term.
We’ll solve for the variable in a radical equation by isolating the radical, squaring both sides and then using inverse operations.
Example
Solve for the variable.
???2x+\sqrt{x+1}=8???
We’ll get the square root by itself so that we can square both sides in order to get rid of the square root.
???2x+\sqrt{x+1}=8???
???2x-2x+\sqrt{x+1}=8-2x???
???\sqrt{x+1}=8-2x???
Squaring both sides, we get
???\left(\sqrt{x+1}\right)^2=(8-2x)^2???
???x+1=64-32x+4x^2???
Now let’s get all of the terms to one side of the equation.
???x-x+1-1=64-1-32x-x+4x^2???
???0=63-33x+4x^2???
Factor the equation and solve for ???x???.
???0=(3-x)(21-4x)???
???x=3??? or ???\frac{21}{4}???
We have to test both roots in the original equation to make sure they’re both solutions. If we plug in ???x=3??? we get
???2(3)+\sqrt{3+1}=8???
???6+\sqrt{4}=8???
???6+2=8???
???8=8???
If we plug in ???x=21/4??? we get
???2\left(\frac{21}{4}\right)+\sqrt{\frac{21}{4}+1}=8???
???\frac{21}{2}+\sqrt{\frac{21}{4}+\frac44}=8???
???\frac{21}{2}+\sqrt{\frac{25}{4}}=8???
???\frac{21}{2}+\frac{5}{2}=8???
???\frac{26}{2}=8???
???13=8???
Because ???8=8??? is a true equation, ???x=3??? is a valid solution. But since ???13=8??? isn’t a true equation, ???x=21/4??? isn’t a valid solution. So the only value that satisfies the equation is
???x=3???
