Content review: The Real Number Mystery: Radical Investigations

✓ Correct: Correct. √49 = 7, and 7 can be written as 7/1.

✗ If B: 49 is a perfect square (7 × 7), so √49 = 7 = 7/1, which is rational.

✓ Correct: Correct. √18 = 3√2, which is irrational.

✗ If A: 18 is not a perfect square. √18 = 3√2, and √2 cannot be written as a fraction.

✓ Correct: Correct. A repeating decimal can be expressed as a fraction. Here, 0.363636… = 4/11.

✗ If B: Not every never-ending decimal is irrational. A repeating decimal is a fraction: 0.363636… = 36/99 = 4/11.

✓ Correct: Correct. ∛(−8) = −2.

✗ If B: (−2) × (−2) × (−2) = −8, so ∛(−8) = −2, an integer. Negative numbers have real cube roots.

✓ Correct: Correct. √72 = √(2 × 2 × 2 × 3 × 3). One pair of 2s and one pair of 3s each send one factor out; one 2 stays inside: 2 × 3 × √2 = 6√2.

✗ If A: Only identical factors form a pair. 72 has a pair of 2s and a pair of 3s; both come out, and only one 2 is left inside.

✗ If D: 8 = 2 × 2 × 2 still contains a pair of 2s. Take that pair out too: 3 × 2 × √2 = 6√2.

✗ If C: A pair brings out one factor, not both. The pairs give 2 × 3 = 6 outside, not 36.

✓ Correct: Correct. √50 = √(2 × 5 × 5). The pair of 5s sends one 5 out; the 2 stays inside: 5√2.

✗ If B: The pair is the two 5s, so 5 comes out. The single 2 has no partner, so it stays inside.

✗ If C: A pair of 5s brings out one 5, not 5 × 5.

✗ If D: The 2 has no partner, so it cannot leave the radical. Only the pair of 5s produces a factor outside.

✓ Correct: Correct. ∛54 = ∛(2 × 3 × 3 × 3). The group of three 3s sends one 3 out; the 2 stays inside: 3∛2.

✗ If C: The complete group is the three 3s, so 3 comes out. The single 2 stays inside.

✗ If A: Cube roots need groups of three identical factors, not pairs. 3 × 3 × 3 is one full group; only the 2 is left inside.

✗ If D: A group of three 3s brings out one 3, not 27.

✓ Correct: Correct. Each complete group of identical factors (a pair for √, three for ∛) is replaced by one factor outside.

✗ If A: Single factors stay inside. In √72, the unmatched 2 had to stay under the radical.

✗ If C: Evenness doesn't matter. In ∛54, a 3 came out because there were three 3s.

✗ If D: The factors are multiplied, not added. A full group of identical factors produces one factor outside.

✓ Correct: Correct. ∛(−16) = −∛(2 × 2 × 2 × 2). One group of three 2s sends one 2 out and one 2 stays inside: −2∛2. Cube roots of negative numbers are real and negative.

✗ If A: Check the sign. ∛(−16) = −∛16, because a negative number times itself three times stays negative.

✗ If C: 16 = 2 × 2 × 2 × 2 has only one complete group of three 2s. The fourth 2 stays inside, so the answer is not a whole number.

✗ If D: Cube roots of negative numbers are real. For example, (−2)³ = −8.

✓ Correct: Correct. The radical parts match, so add the coefficients: 3 + 5 = 8, giving 8√2.

✗ If B: Add the coefficients only. The radical part √2 stays the same, just as 3x + 5x = 8x.

✗ If C: Adding like radicals adds the coefficients: 3 + 5 = 8, not 3 × 5.

✗ If D: The √2 does not disappear: 3√2 + 5√2 = 8√2.

✓ Correct: Correct. The radical parts match, so subtract the coefficients: 7 − 2 = 5, giving 5√3.

✗ If A: This is subtraction: 7 − 2 = 5.

✗ If C: Don't subtract the radicands. √3 is the shared part, just as 7y − 2y = 5y.

✗ If D: Subtract the coefficients and keep √3: 5√3.

✓ Correct: Correct. √8 = 2√2 and √18 = 3√2, so √8 + √18 = 2√2 + 3√2 = 5√2.

✗ If B: √a + √b is not √(a + b). Simplify each radical first: √8 = 2√2 and √18 = 3√2.

✗ If A: Add the coefficients: 2√2 + 3√2 = 5√2. The 2 and 3 are added, not multiplied.

✗ If D: They look unlike, but simplify first: √8 = 2√2 and √18 = 3√2 are like radicals.

✓ Correct: Correct. ∛16 = 2∛2 and ∛54 = 3∛2, so ∛16 + ∛54 = 2∛2 + 3∛2 = 5∛2.

✗ If A: Cube roots can't be added by adding the radicands. Simplify first: ∛16 = 2∛2 and ∛54 = 3∛2.

✗ If B: ∛54 = 3∛2, not 3∛3. Unlike radicals such as ∛2 and ∛3 can never be combined into one radical.

✗ If D: These are cube roots, so the answer keeps the cube root: 5∛2.

✓ Correct: Correct. √3 × √12 = √36 = 6.

✗ If A: Multiply the radicands, don't add them: √3 × √12 = √(3 × 12) = √36.

✗ If B: 36 is a perfect square. Simplify: √36 = 6.

✗ If D: √3 × √12 = √36, and √36 = 6, not 36.

✓ Correct: Correct. 2 × 3 × 5 = 30, because √5 × √5 = 5.

✗ If A: √5 × √5 = 5, so the product is 6 × 5.

✗ If D: Multiply the coefficients (2 × 3 = 6) and the radicals (√5 × √5 = 5), instead of adding them.

✗ If C: √25 is a perfect square: 6√25 = 6 × 5 = 30.

✓ Correct: Correct. √2 × √3 = √6. No perfect-square factor can be extracted.

✗ If B: Multiply the radicands, don't add them: √(2 × 3) = √6.

✗ If A: √2 × √3 = √(2 × 3) = √6, not 2 × 3.

✗ If D: Square roots can always be multiplied: √a × √b = √(ab) when a, b ≥ 0.

✓ Correct: Correct. ∛4 × ∛16 = ∛64 = 4, because 4 × 4 × 4 = 64.

✗ If A: Multiply the radicands, don't add them: ∛(4 × 16) = ∛64.

✗ If B: Multiply inside the cube root: 4 × 16 = 64. The cube root of 64 is 4 because 4 × 4 × 4 = 64.

✗ If C: Two irrational cube roots can multiply to a rational number: ∛64 = 4.

✓ Correct: Correct. Sometimes. Example: √2 + √3 is irrational. Counterexample: √2 + (−√2) = 0, which is rational.

✗ If A: √2 + (−√2) = 0, which is rational, so it is not always irrational.

✗ If C: √2 + √3 is irrational, so it can happen.

✓ Correct: Correct. Sometimes. Example: √3 × √3 = 3 is rational. Counterexample: √2 × √3 = √6 is irrational.

✗ If A: √2 × √3 = √6 is irrational, so it is not always rational.

✗ If C: √3 × √3 = 3 is rational, so it can happen.

✓ Correct: Correct. Always. Example: 4 + √2 is irrational. If r + x were rational, then x = (r + x) − r would be a difference of two rational numbers, which is rational. That contradicts x being irrational.

✗ If B: Try to find a rational result. If r + x were rational, then x = (r + x) − r would be rational too. That's impossible, so the sum is always irrational.

✗ If C: 4 + √2 is irrational, so it does happen. In fact it always happens.

✓ Correct: Correct. Always. Example: 3√2 is irrational. If r · x were rational with r ≠ 0, then x = (r · x) ÷ r would be rational. “Nonzero” matters because 0 · x = 0 is rational.

✗ If B: The only exception would be 0 · x = 0, and the word “nonzero” rules that out. With r ≠ 0, x = (r · x) ÷ r, so r · x must be irrational.

✗ If C: 3√2 is irrational, so it does happen. In fact, with a nonzero rational factor it always happens.