Radicals and roots — ACT practice questions

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Practice 41 Radicals and roots questions in the app

What this topic is

Working with square roots depends on a few dependable moves in Radicals and roots.

Multiplying roots combines what is under the sign, so the square root of three times the square root of twelve equals the square root of thirty-six, which is six. Simplifying pulls out perfect squares, turning the square root of twelve into two times the square root of three.

One frequent error is assuming that the root of a sum equals the sum of the roots, which is false. ACT Mathematics presents these as short expressions to simplify, with answer choices that look alike but differ in the number outside or inside the root.

Sample questions

Pick an answer to see whether it is right — nothing is saved, and nothing needs an account.

Question 1Easier

$\sqrt{50} \cdot \sqrt{18} =$ ?

\sqrt{68}

✗ Not this one

15 \sqrt{2}

✗ Not this one30✓ Correct900✗ Not this one

Show answer

Answer: C — 30

$\sqrt{50} = 5 \sqrt{2}$ and $\sqrt{18} = 3 \sqrt{2}$ , so the product is $15 \cdot 2 = 30$ .

Question 2Easier

$(\sqrt{7} + \sqrt{2}) (\sqrt{7} - \sqrt{2}) =$ ?

9 - 2 \sqrt{14}

✗ Not this one

\sqrt{5}

✗ Not this one5✓ Correct9✗ Not this one

Show answer

Answer: C — 5

The factors are conjugates, so their product is the difference of squares: $(\sqrt{7})^{2} - (\sqrt{2})^{2} = 7 - 2 = 5$ .

Question 3Mid

$(2 \sqrt{3} - \sqrt{2})^{2} =$ ?

14 - 4 \sqrt{6}

✓ Correct

14 - 2 \sqrt{6}

✗ Not this one10✗ Not this one

14 + 4 \sqrt{6}

✗ Not this one

Show answer

Answer: A — $14 - 4 \sqrt{6}$

Using $(a - b)^{2} = a^{2} - 2 a b + b^{2}$ gives $x = 12 - 4 \sqrt{6} + 2 = 14 - 4 \sqrt{6}$ .

Every question is picked for where you are right now, wrong answers come back until they stick, and it all works offline.

Practice 41 Radicals and roots questions in the app

Radicals and roots — ACT practice questions

What this topic is

Sample questions

More ACT topics

Mathematics