How to solve equations with radicals

 
 
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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.


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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???

 
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