NBT Mathematics: Exponents, Surds and Equations 4

Question 1

Solve for x: 2^(2x+1) − 5·2^x + 2 = 0

Accepted Answer

x = −1 or x = 1

Answer

Let k = 2^x, so 2^(2x+1) = 2·(2^x)² = 2k². The equation becomes 2k² − 5k + 2 = 0, which factors as (2k − 1)(k − 2) = 0. So k = 1/2 → 2^x = 2^(−1) → x = −1, or k = 2 → 2^x = 2^1 → x = 1. Both solutions are valid.

Question 2

If 5^p = 3, what is the value of 5^(2p − 1)?

Accepted Answer

9/5

Answer

Rewrite 5^(2p − 1) = 5^(2p) / 5^1 = (5^p)² / 5. Since 5^p = 3, this becomes 3² / 5 = 9/5.

Question 3

Simplify: (√6 + √2)(√6 − √2) / (√3 − 1)

Accepted Answer

2√3 + 2

Answer

The numerator is a difference of squares: (√6)² − (√2)² = 6 − 2 = 4. So the expression becomes 4/(√3 − 1). Rationalizing by multiplying by (√3 + 1)/(√3 + 1) gives 4(√3 + 1)/(3 − 1) = 4(√3 + 1)/2 = 2√3 + 2.

Question 4

Solve √(3x + 4) = x − 2, and state only the solution(s) that satisfy the original equation.

Accepted Answer

x = 7

Answer

Squaring both sides: 3x + 4 = x² − 4x + 4, giving x² − 7x = 0, so x(x − 7) = 0. This yields x = 0 or x = 7. However, x = 0 requires the right-hand side x − 2 = −2, which is negative — impossible for a square root. Only x = 7 is valid: √(25) = 5 = 7 − 2 ✓.

Question 5

Simplify: (27^(2/3) × 16^(−3/4)) / (8^(1/3) × 9^(−1/2))

Accepted Answer

27/16

Answer

Evaluate each factor: 27^(2/3) = (3³)^(2/3) = 3² = 9; 16^(−3/4) = (2⁴)^(−3/4) = 2^(−3) = 1/8; 8^(1/3) = (2³)^(1/3) = 2; 9^(−1/2) = (3²)^(−1/2) = 1/3. So the expression = (9 × 1/8) / (2 × 1/3) = (9/8) ÷ (2/3) = (9/8) × (3/2) = 27/16.

(NBT) National Benchmark Test Practice Test

(NBT) National Benchmark Test Practice Test

NBT Mathematics: Exponents, Surds and Equations 4