ACT Math Quiz: Radical And Exponential Functions

What this quiz covers

This quiz focuses on Radical And Exponential Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACT Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 10
2. 40 (correct answer)
3. 89
4. 1,600

Explanation: This is a formula evaluation question testing radical simplification. Choice B (40) is correct — substitute d = 25: v = √(64 × 25) = √1600 = 40. A faster approach: √(64 × 25) = √64 × √25 = 8 × 5 = 40. Choice A (10) likely comes from computing √64 + √25 = 8 + 5... wait, that's 13. A (10) more likely: student divides instead of multiplying: √(64/25) ≈ 1.6... or computes 64 − 25 = 39, √39 ≈ 6.2. Most likely: student evaluates √64 = 8 and divides by √(25/something). Choice C (89) comes from adding before taking the root: √(64 + 25) = √89 ≈ 9.4, then rounds up to 89 or misreads. Choice D (1,600) correctly multiplies 64 × 25 = 1,600 but forgets to take the square root. Pro tip: When a formula contains a square root, evaluate everything inside the radical first, then take the root. You can also split √(ab) = √a × √b when both are perfect squares — a useful shortcut here since both 64 and 25 are perfect squares.

Question 2

1. 2,400
2. 3,600
3. 10,800 (correct answer)
4. 32,400

Explanation: This is an exponential models question testing careful evaluation of a fractional exponent. Choice C (10,800) is correct — substitute t = 6: P(6) = 400(3)^(6/2) = 400(3)³ = 400 × 27 = 10,800. Choice A (2,400) treats the exponent as a multiplier: 400 × 6 = 2,400, ignoring the exponential structure entirely. Choice B (3,600) uses exponent 2 instead of 3: 400 × 3² = 400 × 9 = 3,600 — computing 6/2 as 2 rather than 3, or using n − 1 = 2 from sequence thinking. Choice D (32,400) uses exponent 4 instead of 3: 400 × 3⁴ = 400 × 81 = 32,400 — perhaps computing (6/2) + 1 = 4. Pro tip: Always resolve the exponent completely before computing the power. Here, t/2 = 6/2 = 3, so the base 3 is raised to the 3rd power: 3³ = 27. Writing it out as 400 × 3 × 3 × 3 = 400 × 27 avoids confusion about what the exponent is.

Question 3

1. $4$
2. $\frac{32}{3}$
3. $64$ (correct answer)
4. $256$

Explanation: The correct answer is C (64). To solve x^(2/3) = 16, raise both sides to the power 3/2 (the reciprocal of 2/3): x = 16^(3/2). Evaluate: 16^(3/2) = (√16)³ = 4³ = 64. A (4) comes from applying only the square root part: √16 = 4, stopping there without cubing. D (256) results from squaring 16 instead of applying the 3/2 power: 16² = 256. B (32/3) treats the fractional exponent as multiplication: 16 × (2/3) = 32/3. Pro tip: to undo x^(m/n) = k, raise both sides to the power n/m — the reciprocal of the exponent.

Question 4

1. $x$ is an integer.
2. $x$ is a rational number.
3. $x$ is a positive real number.
4. $x$ is a nonnegative real number. (correct answer)

Explanation: This is a domain of expressions question testing the restriction on even-index radicals. Choice D (x is a nonnegative real number) is correct — for ⁶√x to produce a real number y, the radicand x must be ≥ 0. This is because even-index roots of negative numbers are not real. x = 0 gives y = 0, which is valid. So x ≥ 0, meaning x is nonnegative. Choice A (x is an integer) is too restrictive — x = 0.5 gives ⁶√0.5, which is a perfectly valid real number. Choice B (x is rational) is also too restrictive — x = √2 (irrational) gives a valid real 6th root. Choice C (x is a positive real number) excludes x = 0, which produces ⁶√0 = 0, a valid real output. Pro tip: The key constraint for real nth roots: when n is even (2, 4, 6, ...), the radicand must be ≥ 0. When n is odd (3, 5, 7, ...), the radicand can be any real number. Here n = 6 (even), so x ≥ 0. Zero must be included because ⁶√0 = 0 is real and valid.

Question 5

1. $32$
2. $16$
3. $64$ (correct answer)
4. $12$

Explanation: To evaluate $4^3$, we multiply 4 by itself three times. $4^3 = 4 \times 4 \times 4 = 16 \times 4 = 64$. The exponential notation means repeated multiplication of the base. Choice B ($16$) would be $4^2$, not $4^3$.

Question 6

1. $x \le 5$
2. $x \ge 5$ (correct answer)
3. $x > 0$
4. All real numbers

Explanation: The correct answer is B (x ≥ 5). For a square root to produce a real value, the expression under the radical must be non-negative: x − 5 ≥ 0 → x ≥ 5. A (x ≤ 5) correctly identifies 5 as the boundary but flips the direction — thinking the square root limits x to values below 5. C (x > 0) applies a general positivity condition without accounting for the −5 shift in the radicand. D (all real numbers) ignores the square root restriction entirely. Pro tip: set the expression inside the square root greater than or equal to zero, then solve that inequality.

Question 7

1. 40.03
2. 400.03
3. 4,000.03 (correct answer)
4. 40,000.03

Explanation: Scientific notation problems require understanding place value and how powers of 10 work. 4 × 10³ = 4,000 and 3 × 10⁻² = 0.03. Adding these: 4,000 + 0.03 = 4,000.03. The key insight is recognizing that 10³ affects the 4 and 10⁻² affects the 3, but they're in completely different place values so they simply add. Don't let the scientific notation intimidate you—convert to standard form, add, and you're done.

Question 8

1. $x \ge 0$
2. $x \ge 5$ (correct answer)
3. $x \le 5$
4. All real numbers

Explanation: Domain problems for radical functions require understanding that even-indexed roots (square roots, fourth roots, etc.) cannot have negative radicands in the real number system. For f(x) = ⁴√(x - 5), the expression under the radical must be non-negative: x - 5 ≥ 0, so x ≥ 5. Note that even roots allow zero but not negative values (unlike odd roots, which allow any real number). This question specifically states "in the real number system" to clarify that complex numbers are not considered.

Question 9

1. 0
2. 2 (correct answer)
3. 3
4. 4

Explanation: This is an exponents and integer reasoning question. Choice B (2) is correct — since 3^x · 3^y = 3^(x+y), and 81 = 3⁴, we have x + y = 4. Both x and y must be positive integers, so the valid pairs are: (1, 3), (2, 2), and (3, 1). The corresponding values of x − y are −2, 0, and 2. The greatest is 2, achieved when (x, y) = (3, 1). Choice A (0) only considers the symmetric case (x, y) = (2, 2), ignoring the other valid pairs. Choice C (3) would require the pair (3.5, 0.5) or (4, 1) — neither consists of positive integers that sum to 4. Choice D (4) would require the pair (4, 0) — but y must be a positive integer (y > 0), so y = 0 is excluded. Pro tip: When a problem says "positive integers," remember that zero does NOT count. List all valid integer pairs that satisfy the constraint, compute the target expression for each, and identify the maximum.

Question 10

1. 2 (correct answer)
2. 4
3. 7
4. 13

Explanation: This is a radical equations question testing the squaring technique. Choice A (2) is correct — square both sides: (√(2x + 5))² = 3² → 2x + 5 = 9. Subtract 5: 2x = 4. Divide by 2: x = 2. Check: √(2(2) + 5) = √9 = 3 ✓. Choice B (4) results from solving 2x = 9 − 1 = 8 → x = 4, subtracting 1 from 9 instead of 5. Choice C (7) comes from adding 5 to 9 instead of subtracting: 2x = 9 + 5 = 14 → x = 7 — flipping the sign when moving 5 to the right side. Choice D (13) results from a double-squaring error: squaring the 3 to get 9, then squaring again before solving, or treating √(2x + 5) = 9 and solving 2x + 5 = 81 → x = 38... Pro tip: After squaring both sides of a radical equation, solve the resulting linear equation normally. Always check your answer by substituting back — squaring can introduce extraneous solutions.

Question 11

1. $(-2, 3)$
2. $(2, -3)$
3. $(-2, -3)$
4. $(2, 3)$ (correct answer)

Explanation: The correct answer is D (2, 3). The vertex form of an absolute value function is y = |x − h| + k, with vertex at (h, k). In y = |x − 2| + 3, h = 2 and k = 3, so the vertex is (2, 3). A (−2, 3) flips the sign of h — since the expression is (x − 2), the x-value that makes it zero is +2, not −2. B (2, −3) correctly identifies x = 2 but flips the vertical shift, reading +3 as −3. C (−2, −3) flips both signs. Pro tip: the vertex is where the expression inside the absolute value equals zero. Set x − 2 = 0 → x = 2, and the minimum y-value is the constant outside: +3.

Question 12

1. $\frac{3x + 5}{2}$ (correct answer)
2. $\frac{3x - 5}{2}$
3. $\frac{2x + 5}{3}$
4. $\frac{3}{2x - 5}$

Explanation: The correct answer is A ((3x + 5)/2). To find the inverse, replace f(x) with y, then swap x and y and solve for y: y = (2x − 5)/3 → swap: x = (2y − 5)/3 → 3x = 2y − 5 → 2y = 3x + 5 → y = (3x + 5)/2. B ((3x − 5)/2) results from a sign error when moving −5 to the other side: treating 3x = 2y + 5 instead of 2y − 5, giving y = (3x − 5)/2. C ((2x + 5)/3) changes the sign of the constant but keeps the original variable structure — a partial attempt that doesn't properly swap x and y. D (3/(2x − 5)) takes the reciprocal of the original function instead of finding the inverse. The inverse undoes the original; the reciprocal is a different operation entirely.

Question 13

1. $\sqrt{64}$
2. $\sqrt{16}$
3. $8$
4. $4\sqrt{2}$ (correct answer)

Explanation: To simplify $\sqrt{32}$, we factor out perfect squares from under the radical. $\sqrt{32}$ = $\sqrt{16 \times 2}$ = $\sqrt{16} \times \sqrt{2}$ = $4\sqrt{2}$. The simplified form is $4\sqrt{2}$ since 16 is a perfect square. Choice C (8) would be if we incorrectly treated $\sqrt{32}$ as $\sqrt{64}$.

Question 14

1. $\sqrt{24}$
2. $\sqrt{6}$
3. $4$
4. $2\sqrt{3}$ (correct answer)

Explanation: To simplify $\sqrt{12}$, we factor out perfect squares from under the radical. $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}$. The simplified form is $2\sqrt{3}$ since 4 is a perfect square. Choice C ($4$) would be if we incorrectly treated $\sqrt{12}$ as $\sqrt{16}$.

Question 15

1. $-16$
2. $16$ (correct answer)
3. $-8$
4. $8$

Explanation: To evaluate (-2)⁴, we apply the exponent to the entire base (-2). Since the exponent 4 is even, the result is positive: (-2)⁴ = (-2) × (-2) × (-2) × (-2) = 16. An even exponent always produces a positive result when applied to a negative base. Choice A incorrectly gives -16, which would be the result if the negative sign were outside the parentheses.

Question 16

1. $-a^4$
2. $\dfrac{1}{a^4}$ (correct answer)
3. $\dfrac{1}{4a}$
4. $a^4$

Explanation: The expression a^(-4) follows the negative exponent rule: a^(-n) = 1/a^n for a ≠ 0. Therefore, a^(-4) = 1/a⁴. The negative exponent means we take the reciprocal of the base raised to the positive exponent. Choice A incorrectly applies the negative sign to the result rather than using the reciprocal rule.

Question 17

1. $9.8\pi$
2. $2\pi$
3. $4\pi$
4. $\pi$ (correct answer)

Explanation: This problem tests function evaluation with a real-world application—always read the formula carefully and substitute values precisely. Given T = $2\pi\sqrt{\frac{L}{9.8}}$ and L = 2.45, substitute: T = $2\pi\sqrt{\frac{2.45}{9.8}}$ = $2\pi\sqrt{\frac{1}{4}}$ = $2\pi \left(\frac{1}{2}\right)$ = $\pi$. Notice that 2.45/9.8 simplifies to exactly 1/4, which makes the square root clean. On formula-based problems, look for these convenient numerical relationships—the ACT designs them to work out cleanly without a calculator.

Question 18

1. $-12$
2. $1$
3. $12$ (correct answer)
4. $0$

Explanation: The first power rule states that any number raised to the power of 1 equals itself. $12^1 = 12$. This is a fundamental property of exponents where the exponent of 1 leaves the base unchanged. Choice B incorrectly gives 1, confusing the exponent value with the result.

Question 19

1. $6\sqrt{2}$ (correct answer)
2. $\sqrt{144}$
3. $12$
4. $\sqrt{36}$

Explanation: To simplify $√72$, we factor out perfect squares from under the radical. $√72$ = $√(36 × 2)$ = $√36$ × $√2$ = $6√2$. The simplified form is $6√2$ since 36 is a perfect square. Choice C ($12$) would be if we incorrectly treated $√72$ as $√144$.

Question 20

1. $2\sqrt{11}$ (correct answer)
2. $\sqrt{22}$
3. $4$
4. $\sqrt{88}$

Explanation: To simplify $\sqrt{44}$, we factor out perfect squares from under the radical. $\sqrt{44} = \sqrt{(4 \times 11)} = \sqrt{4} \times \sqrt{11} = 2\sqrt{11}$. The simplified form is $2\sqrt{11}$ since 4 is a perfect square. Choice C ($4$) would be if we incorrectly treated $\sqrt{44}$ as $\sqrt{16}$.