Solve radical equations

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Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 27 minutes
Domains: Advanced Math
Techniques: Radical-equationsIsolate-and-powerDomain-restrictionsExtraneous-candidates

What you’ll learn

Spot a one-variable equation where the unknown sits inside a radical, like $\sqrt{x+8}=x-2$ .
Graph the two original sides when the points where they meet give you the answer.
Solve by hand when clearing the radical is short, or when you need an exact answer.
Solve the equation that's left once the radical is gone.
See ahead of time which values can't work, because the expression under a square root can't be negative and neither can the root itself.
Check every answer in the original equation and throw out any that fail.

Prerequisites

Why this matters on the SAT

Clear the radical, then check the original equation

Squaring gets rid of a square root and shows you the simpler equation hiding underneath. But you're not done when you solve that equation. Your answer still has to work in the original, the one with the square root.

Solution to the example

The square root is already by itself on the left, so square both sides:

\begin{aligned} \left(\sqrt{4x+5}\right)^2&=9^2,\\[1.4em] 4x+5&=81,\\[1.4em] 4x&=76,\\[1.4em] x&=19. \end{aligned}

Now put $19$ back into the original equation:

\sqrt{4(19)+5}=\sqrt{81}=9.

It works, so the answer is B. Notice that $76$ is waiting in choice D for anyone who stops one step early.

SAT example

Which choice gives the value of $x$ that satisfies

\sqrt{4x+5}=9?

A
$4$
B
$19$
C
$22$
D
$76$

Choose the shorter way in

When the unknown sits inside a radical, you have two good ways in. Start by noticing any limits the original equation puts on $x$ , like a square root that can't equal a negative number. Then pick whichever way is less work and less likely to go wrong:

Graph it in Desmos when you can read the answer straight off the graph, as a number to enter or a point that picks out one answer choice. Put the left side on one line and the right side on another, and click where they meet.
Solve it by hand when getting the radical alone and clearing it takes only a line or two, as in $\sqrt{4x+5}=9$ , or when the question needs an exact answer rather than a rounded decimal.

Graphing the two original sides has a real advantage: you're graphing the equation you were given, so it can't invent extra answers the way squaring can. Still, make sure the window shows every $x$ that could work, so an intersection doesn't hide off screen. And if Desmos shows a rounded decimal, confirm the exact value with algebra or by plugging it into the original equation.

Example: Graph first, then reject an extra answer

Worked example

What value of $x$ satisfies the equation

\sqrt{2x+15}=x?

Step 1

Spot the limit on x, then graph both sides

Look at the right side. It equals $\sqrt{2x+15}$ , and a square root is never negative, so any solution has to satisfy

x\ge0.

Both sides are quick to type in:

y=sqrt(2x+15)
y=x

The graphs meet only at $x=5$ . The question asks for a number, so on test day you'd enter $5$ and move on. The same goes for a multiple-choice question whose choices this value picks out.

Step 2

Square both sides and watch an extra answer appear

Let's keep going anyway, because the algebra shows what squaring does. Square both sides, move every term to one side, and factor:

\begin{aligned} \left(\sqrt{2x+15}\right)^2&=x^2,\\[1.4em] 2x+15&=x^2,\\[1.4em] x^2-2x-15&=0,\\[1.4em] (x-5)(x+3)&=0. \end{aligned}

The squared equation gives two possible answers, $x=5$ and $x=-3$ . But the graph showed only one.

Step 3

Check both in the original equation

Put each value into the equation from the problem, not the squared one:

\begin{aligned} x=5:\quad &\sqrt{2(5)+15}=\sqrt{25}=5,\\[1.4em] x=-3:\quad &\sqrt{2(-3)+15}=\sqrt9=3\ne-3. \end{aligned}

The value $5$ works. The value $-3$ gives $3$ on the left and $-3$ on the right, so it fails. Squaring added it, and the graph had it right all along:

\boxed{x=5}.

You could have seen this coming. The value $-3$ breaks the limit $x\ge0$ from the first step, and a square root can't equal a negative number.

Common mistake:

It’s tempting to check your answers in the squared equation, $2x+15=x^2$ . But both $5$ and $-3$ make that equation true, because squaring erased the sign that tells them apart. Only the original, $\sqrt{2x+15}=x$ , can catch the extra answer.

Calculator loads as you approach

The two original sides meet only at x = 5. Because you graphed the original equation, the extra value x = -3 never shows up.

Isolate before you use a power

When solving by hand is shorter, get the radical alone on one side before you square. Take

\sqrt{3x+10}+4=11.

Subtract $4$ first. Then square both sides:

\begin{aligned} \sqrt{3x+10}&=7,\\[1.4em] \left(\sqrt{3x+10}\right)^2&=7^2,\\[1.4em] 3x+10&=49,\\[1.4em] x&=13. \end{aligned}

Check it in the original equation:

\sqrt{3(13)+10}+4=\sqrt{49}+4=11.

What if you square before getting the radical alone? The left side becomes $\left(\sqrt{3x+10}+4\right)^2$ , and expanding it gives a middle term, $8\sqrt{3x+10}$ . The square root is still there, so squaring made the equation longer instead of clearing the radical.

Common mistake:

A common slip is squaring each term on its own, like turning $\sqrt{3x+10}+4=11$ into $3x+10+16=121$ . That leads to $x=\frac{95}{3}$ instead of $13$ . When you square an equation, you square the whole left side and the whole right side. So get the radical alone, picture parentheses around each side, and then raise both sides to the matching power.

Why squaring can create an extra answer

An extraneous solution is a value that solves the squared equation, or any other changed version, but doesn't solve the original. Squaring can create one because a number and its opposite have the same square:

3^2=(-3)^2=9.

After squaring, the equation can no longer tell $3$ from $-3$ .

Here's the general picture. Say the equation looks like $\sqrt{R(x)}=L(x)$ , with a square root on one side and some expression $L(x)$ on the other. A true solution has to pass two tests:

R(x)\ge0 \quad\text{and}\quad L(x)\ge0.

The first says you can't take the square root of a negative number. The second says $L(x)$ equals a square root, so it can't be negative either. The squared equation drops that second test, so it may accept a value that makes $L(x)<0$ . That value solves the squared equation, not the original.

Check your understanding:

Before solving $\sqrt{x+8}=x-2$ , what two limits on $x$ can you read from the original equation?

Common mistake:

It’s a common slip to write $\sqrt{25}=\pm5$ . A plus-or-minus pair comes from solving an equation like $u^2=25$ , where $u$ could be $5$ or $-5$ . The symbol $\sqrt{25}$ names one number, the principal root, so $\sqrt{25}=5$ .

Match the power to the root

You clear a square root by squaring. Other roots work the same way: for an $n$ th root that's alone on one side, raise both sides to the $n$ th power.

\sqrt[n]{R(x)}=L(x) \quad\longrightarrow\quad R(x)=\left(L(x)\right)^n.

So for a cube root, cube both sides. For example,

\sqrt[3]{2x-5}=3

becomes

\begin{aligned} 2x-5&=27,\\[1.4em] x&=16. \end{aligned}

Check it:

\sqrt[3]{2(16)-5}=\sqrt[3]{27}=3.

Cubing is different from squaring in one useful way. Two different real numbers never have the same cube: $3^3=27$ , but $(-3)^3=-27$ . In math terms, the cube function is one-to-one over the real numbers. So cubing both sides of a real equation doesn't lose any sign information. Checking is still a good last habit, especially after several algebra steps.

Check your understanding:

What power clears $\sqrt[4]{5x+1}=2$ , and what value of $x$ does it give?

Practice problems

Before each problem, look for limits on $x$ , then decide whether to graph the two sides or clear the radical by hand.

Clear a cube root

Practice problem

What value of $x$ satisfies

\sqrt[3]{5x-7}=3?

Your answer

Calculator loads as you approach

By hand: cubing both sides is quickest. Use the calculator only if you want to check your answer.

Graph, then reject an extra answer

Practice problem

Which choice is the solution to

\sqrt{3x+4}+2=x?

Calculator loads as you approach

Graph it: put each original side on its own line and click where they meet.

Use two radicals to reach an expression directly

Practice problem

\sqrt{x+8}+\sqrt{17-x}=7,

what is the value of $x^2-9x$ ?

Your answer

Calculator loads as you approach

By hand. Keep the expression the question asks about in view, because the algebra reaches it before you find x.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Look for limits on $x$ first, then pick the way with less work and less risk.
Graph the two original sides when the point where they meet gives a number or picks out an answer choice.
By hand, get the radical alone, then raise both whole sides to the matching power.
A value you find after squaring is only a possible answer until it works in the original equation.
Keep what the question asks for in view. You may not need to find each solution.

Next lesson

Solve rational equations

Track denominator restrictions, clear fractions, and reject values that make an original denominator zero.

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Practice

Practice this lesson

148 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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