These eleven questions mix the five skills in this module: quadratic inequalities, modulus equations, surd equations, exponential equations and algebraic fractions. They run from straightforward to harder. Work each one on paper, then open the answer and compare your method, not just your final value.
Keep a note of each slip you make. The mistake log and retest queue is one place to record it, and the non-calculator working trainer is useful for practising exact checks. Return to the module overview if you want the suggested study order.
Questions
Q1. Solve |x − 4| = 6.
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Case 1: x − 4 = 6, so x = 10. Case 2: x − 4 = −6, so x = −2.
Check: |10 − 4| = 6 and |−2 − 4| = 6.
x = 10 or x = −2
Q2. Solve x² − 7x + 10 < 0.
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x² − 7x + 10 = (x − 2)(x − 5), so the roots are 2 and 5. Testing x = 3 gives (1)(−2) = −2, which is negative. The expression is negative only between the roots.
2 < x < 5
Q3. Solve 2^(x − 1) = 16.
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16 = 2⁴, so x − 1 = 4 and x = 5. Check: 2⁴ = 16.
x = 5
Q4. Solve x² + x − 12 ≥ 0.
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x² + x − 12 = (x + 4)(x − 3), so the roots are −4 and 3. Testing x = 0 gives −12, so the middle region is negative. The required regions are the two outer ones, and the roots are included.
x ≤ −4 or x ≥ 3
Q5. Solve |3x + 2| = x + 6.
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Case 1: 3x + 2 = x + 6, so 2x = 4 and x = 2. Case 2: 3x + 2 = −(x + 6) = −x − 6, so 4x = −8 and x = −2.
Check x = 2: |8| = 8 and 2 + 6 = 8. Check x = −2: |−4| = 4 and −2 + 6 = 4.
x = 2 or x = −2
Q6. Solve √(2x + 8) = x.
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Square: 2x + 8 = x², so x² − 2x − 8 = 0, which gives (x − 4)(x + 2) = 0. Candidates: 4 and −2.
Check 4: √16 = 4, matches. Check −2: √4 = 2, but the right side is −2, so it fails.
x = 4
Q7. Solve 4^(x + 1) = 32^x.
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Base 2: 4^(x + 1) = 2^(2x + 2) and 32^x = 2^(5x). So 2x + 2 = 5x, giving 3x = 2 and x = 2/3.
Check: 4^(5/3) = 2^(10/3) and 32^(2/3) = 2^(10/3).
x = 2/3
Q8. State the restriction, then solve 6/(x − 2) = x − 1.
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Restriction: x ≠ 2. Multiply by (x − 2): 6 = (x − 1)(x − 2) = x² − 3x + 2, so x² − 3x − 4 = 0, which gives (x − 4)(x + 1) = 0.
Neither root is 2. Check x = 4: 6/2 = 3 and 4 − 1 = 3. Check x = −1: 6/(−3) = −2 and −1 − 1 = −2.
x = 4 or x = −1
Q9. Solve √(x + 10) = x − 2.
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Square: x + 10 = x² − 4x + 4, so x² − 5x − 6 = 0, which gives (x − 6)(x + 1) = 0. Candidates: 6 and −1.
Check 6: √16 = 4 and 6 − 2 = 4, matches. Check −1: √9 = 3, but −1 − 2 = −3, so it fails.
x = 6
Q10. Solve 3x² ≤ 5x + 2.
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Rewrite: 3x² − 5x − 2 ≤ 0. Factorise: (3x + 1)(x − 2) ≤ 0, so the roots are −1/3 and 2. Testing x = 0 gives −2, which is negative, so the middle region satisfies the inequality and the roots are included.
−1/3 ≤ x ≤ 2
Q11. Solve 2^(2x) − 6(2^x) + 8 = 0.
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Let y = 2^x. Then y² − 6y + 8 = 0, so (y − 2)(y − 4) = 0 and y = 2 or y = 4.
2^x = 2 gives x = 1, and 2^x = 4 gives x = 2. Check x = 1: 4 − 12 + 8 = 0. Check x = 2: 16 − 24 + 8 = 0.
x = 1 or x = 2
If you got these wrong
Find the error type below, revise the lesson, then attempt a fresh question of the same kind.
- Wrote one answer for a modulus equation, or only negated part of the right side (Q1, Q5): revisit solving an equation involving an absolute value.
- Gave a two-piece inequality answer as one chain, or chose the wrong region (Q2, Q4, Q10): revisit solving a quadratic inequality using sign intervals. The quadratic structure explorer shows the curve and its regions.
- Kept a root that fails the original equation (Q6, Q9): revisit rejecting an extraneous root after squaring.
- Compared the exponents without first making the bases the same, or could not rewrite a base (Q3, Q7, Q11): revisit solving an exponential equation by a common base.
- Forgot a restriction or missed a zero denominator (Q8): revisit stating restrictions before manipulating fractions.
If a pattern of slips stays after revising the lessons, it may help to talk through your working with someone who can see it as you write it. That is what online one-to-one Additional Mathematics tuition is designed for.