Practice Test Geeks(NBT) National Benchmark Test Practice Test
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NBT Mathematics: Exponents, Surds and Equations Flashcards

6 cards from real NBT practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.

Read the first 6 NBT Mathematics: Exponents, Surds and Equations flashcards as text

  1. Simplify the expression: (2^(x+1) · 4^(x−1)) / 8^(x−1)

    Answer: 4

    Convert all bases to powers of 2: 4^(x−1) = 2^(2x−2) and 8^(x−1) = 2^(3x−3). The numerator becomes 2^(x+1) · 2^(2x−2) = 2^(3x−1). The denominator is 2^(3x−3). Subtracting exponents: 2^(3x−1−(3x−3)) = 2^(3x−1−3x+3) = 2^2 = 4. The x terms cancel completely, making the result a constant.

  2. Solve for x: 9^x − 12 · 3^x + 27 = 0

    Answer: x = 1 or x = 2

    Recognise that 9^x = (3^2)^x = (3^x)^2. Let k = 3^x, giving the quadratic k² − 12k + 27 = 0. Factorising: (k − 3)(k − 9) = 0, so k = 3 or k = 9. Solving 3^x = 3 gives x = 1, and 3^x = 9 = 3^2 gives x = 2. A common error is treating k = 3 and k = 9 as the final answers without converting back.

  3. Solve for x: √(2x − 1) + √(x − 4) = 4, and identify which solutions are valid.

    Answer: x = 5 only

    Isolate one radical: √(2x−1) = 4 − √(x−4). Square both sides: 2x−1 = 16 − 8√(x−4) + (x−4), simplifying to x − 13 = −8√(x−4), or 13 − x = 8√(x−4). Square again: (13−x)² = 64(x−4), giving x² − 90x + 425 = 0, so (x−5)(x−85) = 0. Checking x = 5: √9 + √1 = 3+1 = 4 ✓. Checking x = 85: √169 + √81 = 13+9 = 22 ≠ 4 ✗. The value x = 85 is an extraneous root introduced by squaring.

  4. Rationalise and simplify: (√6 + √2) / (√3 + 1)

    Answer: √2

    Multiply numerator and denominator by the conjugate of the denominator (√3 − 1): Numerator: (√6 + √2)(√3 − 1) = √18 − √6 + √6 − √2 = 3√2 − √2 = 2√2. Denominator: (√3 + 1)(√3 − 1) = 3 − 1 = 2. Result: 2√2 / 2 = √2. Notice that the intermediate √6 terms cancel neatly.

  5. Simplify √(8 − 2√15) into the form √a − √b, where a > b > 0.

    Answer: √5 − √3

    Assume √(8 − 2√15) = √a − √b. Squaring: a + b = 8 and 2√(ab) = 2√15, so ab = 15. Solving a + b = 8 and ab = 15 gives a = 5 and b = 3 (roots of t² − 8t + 15 = 0). Therefore √(8 − 2√15) = √5 − √3. Verify: (√5 − √3)² = 5 − 2√15 + 3 = 8 − 2√15 ✓. Since √5 > √3, the expression is positive.

  6. Solve for x: 2^(2x+1) = 3 · 2^x + 2

    Answer: x = 1

    Rewrite 2^(2x+1) as 2 · (2^x)². Let k = 2^x: 2k² = 3k + 2, giving 2k² − 3k − 2 = 0. Factorising: (2k + 1)(k − 2) = 0, so k = −1/2 or k = 2. Since k = 2^x > 0 for all real x, we discard k = −1/2. Therefore 2^x = 2, giving x = 1. Verify: 2^3 = 8 and 3·2 + 2 = 8 ✓.