SAT Radical Expressions and Rational Exponents on Desmos

A rational exponent is a radical in another form: x^(1/2) is the square root of x and x^(2/3) is the cube root of x squared. To check an equivalent-form question on Desmos, type the original and each choice with parentheses around the fractional exponent, test a value such as x = 2, and keep the choice that matches. For a radical equation, graph both sides and read where they cross.

You can run every check below in the College Board version of Desmos on our Desmos on the SAT page. This guide gives you the rule, two worked examples, three practice problems with answers, and the check that catches extraneous solutions.

The rule: x^(m/n) is the nth root of x to the m

The bottom of the fraction names the root. The top names the power. So x^(m/n) means: take the nth root of x, then raise it to the m. You can also raise first and take the root second, and you get the same number.

Two exponent rules do most of the work on SAT radical questions. When you multiply powers of the same base, add the exponents: x^(3/4) · x^(5/4) = x^2. When you raise a power to a power, multiply them: (x^6)^(2/3) = x^4.

On Desmos, always put parentheses around a fractional exponent, as in x^(2/3). Before you read any value, look at the line and check that the whole 2/3 sits up in the exponent.

College Board allows a calculator on the entire Math section, and Bluebook has a Desmos calculator built in with graphing and scientific options. The digital SAT has no separate calculator section, so you can use Desmos on any radical question.

Worked example 1 (I do): find the equivalent expression

SAT-style question. Which expression is equivalent to (27x^6)^(2/3)?

A) 9x^4   B) 3x^4   C) 18x^4   D) 9x^9

Step 1. Define the original as a function. I type it on line 1 with parentheses around the exponent:

1  f(x) = (27x^6)^(2/3)
2  f(2)

Line 2 evaluates the function at x = 2. The value is 144. I pick 2 because it is small and positive. I skip 0 and 1 on purpose, for reasons you will see in a moment.

Step 2. Evaluate each choice at x = 2 the same way. I give choice A its own function and evaluate it:

3  a(x) = 9x^4
4  a(2)

Then I repeat with b(x) = 3x^4, c(x) = 18x^4 and d(x) = 9x^9. At x = 2 the four choices give 144, 48, 288 and 4,608. Only choice A matches 144, so the answer is A.

Step 3. Check by hand. Split the power over each factor. 27^(2/3) = (∛27)² = 3² = 9, and (x^6)^(2/3) = x^(6 · 2/3) = x^4. So (27x^6)^(2/3) = 9x^4. As a numeric check, 27 · 2^6 = 1,728, and 1,728^(2/3) = 12² = 144.

Each wrong choice comes from one slip. B takes the cube root of 27 but forgets to square it. C multiplies 27 by 2/3. D multiplies 6 by 3/2, which flips the exponent.

Now the reason to skip x = 1. At x = 1, choice A gives 9 and choice D also gives 9, so the test cannot tell them apart. If two choices ever tie, try a second value. You can also graph y = f(x) and y = 9x^4; equivalent expressions draw one curve on top of the other.

Worked example 2 (we do): solve a radical equation and drop the extraneous root

SAT-style question. What value of x satisfies √(3x + 7) = x - 1?

Step 1. Square both sides. 3x + 7 = (x - 1)² = x² - 2x + 1.

Step 2. Set the equation to zero. x² - 5x - 6 = 0.

Step 3. Factor. (x - 6)(x + 1) = 0, so the candidates are x = 6 and x = -1.

Step 4 (your step). Squaring can add a solution that the original equation rejects. Put each candidate back into √(3x + 7) = x - 1 and keep only the one that works.

Check your answer: At x = 6, the left side is √25 = 5 and the right side is 6 - 1 = 5. It works. At x = -1, the left side is √4 = 2 but the right side is -2. A square root is never negative, so x = -1 is extraneous. The answer is x = 6.

Desmos shows the same result as a picture. Type each side as its own graph:

1  y = (3x+7)^(1/2)
2  y = x - 1

Select one of the graphs and gray points of interest appear, including the point where the two graphs cross. Hover over it or select it to read the coordinates: (6, 5). The graphs cross only once, so the extraneous x = -1 never shows up. That is why the graph is a good check after you square both sides by hand.

For more on this skill, work through the free Exponents and Radicals lesson notes.

Your turn (you do): 3 practice problems

1. For x > 0, which expression is equivalent to x^(3/4) · x^(5/4)? A) x^2   B) x^(15/16)   C) x^(1/2)   D) x^(15/4)
2. What value of x satisfies √(2x + 3) = x?
3. If x^(2/3) = 16 and x > 0, what is the value of x?

Answers

1. A. Add the exponents: 3/4 + 5/4 = 8/4 = 2. With x = 4, the original is 4^(3/4) · 4^(5/4) = 16, and only 4^2 also gives 16.
2. x = 3. Squaring gives x² - 2x - 3 = 0, so x = 3 or x = -1. At x = -1 the left side is √1 = 1, not -1, so -1 is extraneous. Graph y = (2x+3)^(1/2) and y = x and they cross only at (3, 3).
3. x = 64. Raise both sides to the 3/2 power: x = 16^(3/2) = 4³ = 64. Check: 64^(2/3) = 4² = 16. The stem says x > 0, so answer 64.

When algebra is faster

College Board's calculator guidance says some Math questions are better done without a calculator, even though one is allowed. Exponent-rule questions often fit that group. Adding 3/4 and 5/4 takes one line, and typing four choices takes longer. Use Desmos when the expression has a coefficient and a power together, like (27x^6)^(2/3), or when you want to confirm a radical equation after squaring.

Mistakes to avoid

• Leaving out the parentheses. Type x^(2/3), not x^2/3. Look at the line on screen and make sure the full fraction sits in the exponent before you trust the value.
• Testing with x = 0 or x = 1. Every power of 1 is 1, so choices tie. In worked example 1, x = 1 gives 9 for both A and D. Use x = 2, or another small positive number.
• Keeping every root after squaring. Squaring both sides of √(3x + 7) = x - 1 gives two candidates, and only x = 6 works. Plug each candidate into the original equation, or check that the graphs cross there.

Practice next

Start with the free Exponents and Radicals lesson notes. For more questions, browse the SAT question bank.

Related guides: for exponent patterns in tables, see SAT exponential functions on Desmos. For equations with higher powers, read how to find all solutions of a polynomial equation. To read the point where two graphs cross, see how to solve a system of equations on Desmos.

Checked on September 26, 2026 against College Board's SAT Suite calculator policy, College Board's March 12, 2024 digital SAT launch release, the College Board version of the Desmos graphing calculator and the Desmos Help Center article Getting Started: Desmos Graphing Calculator. Every answer on this page was recomputed before publishing. The Test Advantage is independent of College Board and Desmos.