A-Level Inequalities Practice Questions

Free A-Level Inequalities practice questions with full step-by-step worked solutions. Covers linear inequality, solve, divide by negative, unknown both sides. Practise exam-style problems and check your method.

linear inequalitysolvedivide by negativeunknown both sidesbracketsquadratic inequality
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Solve the inequality 3x+4>193x + 4 > 19.
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Worked solution

  1. Subtract 4 from both sides

    3x+4−4>19−4  ⇒  3x>153x + 4 - 4 > 19 - 4 \;\Rightarrow\; 3x > 15

    We want the term with x on its own, so we undo the +4 by subtracting 4 from each side. Because we are only adding or subtracting, the inequality sign stays exactly the same.

  2. Divide both sides by 3

    3x3>153  ⇒  x>5\frac{3x}{3} > \frac{15}{3} \;\Rightarrow\; x > 5

    Now we undo the multiplication by 3 by dividing both sides by 3. We are dividing by a positive number, so the inequality sign does not flip.

  3. State the solution

    x>5x > 5

    So every number bigger than 5 makes the inequality true. On the number line we use an open circle at 5 to show 5 itself is not included.

Answer
x>5x > 5
Question 2
2 markseasy
Solve the inequality 4−x<104 - x < 10.
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Worked solution

  1. Subtract 4 from both sides

    −x<6-x < 6

    Take 4 from both sides to leave the x-term by itself.

  2. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  3. Multiply by -1 and flip the sign

    x>−6x > -6

    Multiplying by -1 to make x positive reverses the inequality, so '<' becomes '>'.

Answer
x>−6x > -6
Question 3
4 marksintermediate
Find the set of values of xx for which both x2−9<0x^2 - 9 < 0 and x>0x > 0.
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Worked solution

  1. Solve the quadratic inequality

    x2−9<0⇒(x−3)(x+3)<0⇒−3<x<3x^2 - 9 < 0 \Rightarrow (x-3)(x+3) < 0 \Rightarrow -3 < x < 3

    Factorise the difference of two squares; below the axis is between the roots -3 and 3.

  2. Note the second condition

    x>0x > 0

    The second inequality is already solved: x must be positive.

  3. Find the overlap

    −3<x<3  and  x>0-3 < x < 3 \;\text{and}\; x > 0

    Both conditions must hold at once, so we look for where the two ranges overlap.

  4. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  5. Interpret the final answer

    0<x<30 < x < 3

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  6. State the solution

    0<x<30 < x < 3

    The overlap of 'between -3 and 3' and 'greater than 0' is the values between 0 and 3.

Answer
0<x<30 < x < 3
Question 4
6 markshard
Find the set of values of kk for which x2+6x+k>0x^2 + 6x + k > 0 for all real values of xx.
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Worked solution

  1. Understand the condition

    x2+6x+k>0 for all xx^2 + 6x + k > 0 \text{ for all } x

    An upward parabola stays entirely above the x-axis only when it never touches or crosses it, i.e. it has no real roots.

  2. Recall the no-real-roots condition

    b2−4ac<0b^2 - 4ac < 0

    No real roots means a negative discriminant.

  3. Identify a, b and c

    a=1,b=6,c=ka = 1, \quad b = 6, \quad c = k

    Read the coefficients from x^2 + 6x + k.

  4. Substitute into the discriminant

    62−4(1)(k)<06^2 - 4(1)(k) < 0

    Put the values into b^2 - 4ac < 0.

  5. Simplify

    36−4k<036 - 4k < 0

    Square the 6 to get 36 and 4 times k is 4k.

  6. Subtract 36 from both sides

    −4k<−36-4k < -36

    Move the 36 across.

  7. Divide by -4 and flip the sign

    k>9k > 9

    Dividing by -4 (negative) reverses the inequality from '<' to '>'.

  8. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  9. Interpret the final answer

    k>9k > 9

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  10. State the solution

    k>9k > 9

    For k greater than 9 the parabola never reaches the axis, so the expression is positive for every real x.

Answer
k>9k > 9
Question 5
8 markschallenging
The equation (k+1)x2−2kx+(k−2)=0(k + 1)x^2 - 2kx + (k - 2) = 0 has real roots. Find the set of values of kk.
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Worked solution

  1. Check the degenerate case k = -1

    k=−1:  0⋅x2+2x−3=0⇒x=32k = -1:\; 0\cdot x^2 + 2x - 3 = 0 \Rightarrow x = \tfrac{3}{2}

    If k = -1 the x^2 coefficient is zero and the equation becomes linear, which still has a real root.

  2. Now take k \ne -1 (a genuine quadratic)

    a=k+1,b=−2k,c=k−2a = k + 1, \quad b = -2k, \quad c = k - 2

    For a real quadratic we use the discriminant with these coefficients.

  3. Condition for real roots

    b2−4ac≥0b^2 - 4ac \ge 0

    Real roots (equal or distinct) require a non-negative discriminant.

  4. Substitute

    (−2k)2−4(k+1)(k−2)≥0(-2k)^2 - 4(k + 1)(k - 2) \ge 0

    Put the coefficients into b^2 - 4ac.

  5. Square the first term

    4k2−4(k+1)(k−2)≥04k^2 - 4(k + 1)(k - 2) \ge 0

    (-2k)^2 = 4k^2.

  6. Expand the product

    4k2−4(k2−k−2)≥04k^2 - 4(k^2 - k - 2) \ge 0

    Multiply out (k+1)(k-2) = k^2 - k - 2.

  7. Remove the bracket

    4k2−4k2+4k+8≥04k^2 - 4k^2 + 4k + 8 \ge 0

    Distribute the -4 across the bracket, changing every sign inside.

  8. Simplify

    4k+8≥04k + 8 \ge 0

    The 4k^2 terms cancel, leaving a linear inequality in k.

  9. Solve

    k≥−2k \ge -2

    Subtract 8 and divide by 4 (a positive divisor, so no flip).

  10. Check the degenerate case fits

    k=−1≥−2  ✓k = -1 \ge -2 \; \checkmark

    The linear case k = -1 already satisfies k \ge -2, so it does not need to be added separately.

  11. Recall what an inequality solution looks like

    a range of values, not one number\text{a range of values, not one number}

    Unlike an equation, an inequality usually has a whole range of solutions. We are finding every value that works, not a single answer.

  12. Interpret the final answer

    k≥−2k \ge -2

    The answer describes a set of numbers. Every value in this range makes the original statement true, and every value outside it does not.

  13. Check the boundary points

    closed endpoints included\text{closed endpoints included}

    Because the inequality allows 'equal to', the boundary values are part of the solution and are drawn as solid (filled) circles.

  14. Verify with a test value

    substitute a value from the range\text{substitute a value from the range}

    Pick a number from your solution and put it back into the original inequality. If it gives a true statement, you can be confident the region is correct.

  15. State the solution

    k≥−2k \ge -2

    For every k that is at least -2 the equation has real roots.

Answer
k≥−2k \ge -2

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