Equations & inequalities at A-Level
At GCSE you solved simultaneous equations where both equations are linear, and inequalities that are linear too. A-Level extends both ideas to quadratics: solving a linear equation together with a quadratic one, and solving inequalities where the unknown is squared.
Four skills make up this topic:
- Linear-quadratic simultaneous equations — substituting a linear equation into a quadratic one
- The discriminant and intersections — using it to find how many times a line meets a curve
- Quadratic inequalities — solved using critical values, not by dividing through
- Combining inequalities — finding the set of values that satisfies more than one condition
Linear-quadratic simultaneous equations
Substitute the linear equation into the quadratic one, solve the resulting quadratic for \(x\), then substitute each \(x\)-value back into the linear equation to find the matching \(y\)-value — the linear equation is quicker.
Worked Example 1
Solve the simultaneous equations \(y = 2x + 1\) and \(y = x^2 + x - 5\).
1
Substitute: \(x^2 + x - 5 = 2x + 1\)
2
Rearrange: \(x^2 - x - 6 = 0\)
3
Factorise: \((x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\)
4
Substitute into \(y = 2x + 1\): \(x = 3\) gives \(y = 7\); \(x = -2\) gives \(y = -3\)
Answer\((3, 7)\) and \((-2, -3)\)
Quadratic inequalities
Never divide a quadratic inequality through by \(x\) — you don't know its sign. Instead, rearrange everything to one side, factorise to find the critical values (the roots), then use the shape of the parabola to decide which region satisfies the inequality.
Worked Example 2
Solve \(x^2 - 5x + 6 > 0\).
1
Factorise: \((x - 2)(x - 3) > 0\)
2
Critical values: \(x = 2\) and \(x = 3\)
3
The graph is an upward parabola, so it's positive \((> 0)\) outside the roots
Answer\(x < 2\) or \(x > 3\)
The shading between the curve and the x-axis marks where \(x^2 - 5x + 6 > 0\) — the region between the roots \((2 < x < 3)\) is left unshaded, giving the solution \(x < 2\) or \(x > 3\) found in Worked Example 2.
Worked Example 3
Solve \(2x^2 \geq x + 6\).
1
Rearrange so everything is on one side: \(2x^2 - x - 6 \geq 0\)
2
Factorise: \(ac = -12\), numbers \(-4\) and \(+3\), split and group: \((2x + 3)(x - 2) \geq 0\)
3
Critical values: \(x = -\dfrac{3}{2}\) and \(x = 2\)
4
Upward parabola, \(\geq 0\) outside (and including) the roots
Answer\(x \leq -\dfrac{3}{2}\) or \(x \geq 2\)
The discriminant and intersections
Substituting a line into a curve and looking at the resulting quadratic's discriminant tells you exactly how the line and curve relate: two distinct real roots means two intersection points, a repeated root means the line is a tangent, and no real roots means they never meet.
Worked Example 4
Find the range of values of \(c\) for which the line \(y = x + c\) does not intersect the curve \(y = x^2 + 2x + 5\).
1
Substitute: \(x^2 + 2x + 5 = x + c\)
2
Rearrange: \(x^2 + x + (5 - c) = 0\)
3
No intersection means no real roots, so the discriminant is negative: \(1 - 4(5 - c) < 0\)
4
\(1 - 20 + 4c < 0\), so \(4c < 19\)
Answer\(c < \dfrac{19}{4}\)
Combining inequalities
When a question gives two conditions that must both be true, solve each inequality separately, then find the overlap between the two solution sets — sketching both regions on the same number line makes this easy to see.
Worked Example 5
Find the set of values of \(x\) for which \(x^2 - x - 12 > 0\) and \(x < 3\).
1
Solve the quadratic inequality: \(x^2 - x - 12 = (x - 4)(x + 3) > 0\), so \(x < -3\) or \(x > 4\)
2
Combine with \(x < 3\): the region \(x < -3\) is entirely within \(x < 3\), so it stays
3
The region \(x > 4\) does not overlap with \(x < 3\) at all, so it's rejected
Answer\(x < -3\)
Worked Example 6
The curve \(y = x^2 - 4x + 7\) and the line \(y = 2x - k\) meet at exactly one point. Find the value of \(k\).
1
Substitute: \(x^2 - 4x + 7 = 2x - k\)
2
Rearrange: \(x^2 - 6x + (7 + k) = 0\)
3
Meeting at exactly one point means a repeated root, so the discriminant is zero: \(36 - 4(7 + k) = 0\)
4
\(36 - 28 - 4k = 0\), so \(8 = 4k\)
Answer\(k = 2\)
Exam tips
💡 Tip 1
Never divide a quadratic inequality by \(x\)
Dividing both sides of an inequality by an unknown you can't sign-check is a serious error. Always rearrange to one side and factorise instead.
💡 Tip 2
Sketch the parabola to decide the region
Once you have the critical values, a 5-second mental sketch of the parabola's shape tells you instantly whether the answer is "outside" or "between" the roots — far more reliable than guessing.
💡 Tip 3
Match strict and non-strict inequalities carefully
If the original inequality uses \(\leq\) or \(\geq\), the critical values are included in the answer; if it uses \(<\) or \(>\), they are not. Carry the correct symbol all the way through.
💡 Tip 4
"Meets", "intersects" and "touches" have different meanings
"Touches" or "is a tangent to" means exactly one point (discriminant = 0). "Intersects at two points" means discriminant \(> 0\). "Does not meet" means discriminant \(< 0\). Read the wording carefully before setting up the discriminant condition.
💡 Tip 5
For combined inequalities, always find the overlap
The answer to "A and B" is the intersection of the two regions, not the union — sketch both on one number line and only keep where they overlap.
Common mistakes
Common Mistake 1
Writing the solution the wrong way round
For \(x^2 - 5x + 6 > 0\) (critical values 2 and 3), writing "\(2 < x < 3\)" gives the region between the roots, which is actually where the expression is negative. The correct answer, "outside" the roots, is \(x < 2\) or \(x > 3\).
Common Mistake 2
Dividing a quadratic inequality by \(x\)
For \(x^2 > 3x\), dividing both sides by \(x\) to get \(x > 3\) loses a whole solution branch and is only valid if \(x\) is known to be positive. Rearrange to \(x^2 - 3x > 0\) and factorise instead: \(x(x - 3) > 0\), giving \(x < 0\) or \(x > 3\).
Common Mistake 3
Substituting back into the quadratic equation instead of the linear one
After solving for \(x\) in a linear-quadratic system, substitute each \(x\)-value into the linear equation to find \(y\) — it's faster and avoids re-doing a squaring step. Either equation gives the correct \(y\)-value, but the linear one is far less error-prone.
Common Mistake 4
Taking the union instead of the intersection for combined inequalities
For "\(x^2 - x - 12 > 0\) and \(x < 3\)", combining the two regions by listing everything from both (a union) is wrong — you need the overlap. Check each candidate region against both conditions before including it.
Common Mistake 5
Confusing "no real roots" with "no solutions exist"
A negative discriminant means the line and curve don't intersect — it doesn't mean there's an error in the working. This is a valid, complete answer to "find the values of c for which they do not meet".
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Solve the simultaneous equations \(y = x + 2\) and \(y = x^2 - 4\).Core Skills
Q2Solve the simultaneous equations \(y = 3x - 1\) and \(y = x^2 + x - 1\).Core Skills
Q3Solve \(x^2 - 9 > 0\).Core Skills
Q4Solve \(x^2 - 16 \leq 0\).Core Skills
Q5Solve \(x^2 - 4x > 0\).Core Skills
Exam-Style
Q6Solve the simultaneous equations \(y = x - 1\) and \(y = x^2 - 3x - 1\).Exam-Style
Q7Solve \(2x^2 - 7x - 15 < 0\).Exam-Style
Q8Find the range of values of \(k\) for which \(x^2 + kx + 4 = 0\) has two distinct real roots.Exam-Style
Q9Find the set of values of \(x\) for which \(x^2 + 2x - 3 < 0\) and \(x > -1\).Exam-Style
Q10Find the value of \(c\) for which the line \(y = x + c\) is a tangent to the curve \(y = x^2 - 3x + 5\).Exam-Style
A* Challenge
Q11Find the set of values of \(x\) for which \(x^2 - 5x + 4 > 0\) and \(2x + 1 > 0\).A* Challenge
Q12The line \(y = 2x + k\) intersects the curve \(y = x^2 - x + 3\) at two distinct points. Find the range of values of \(k\).A* Challenge
Q13Find the set of values of \(x\) for which \(x^2 - 2x - 8 \leq 0\), giving your answer in set notation.A* Challenge
Q14Solve the simultaneous equations \(y = x^2 - 6x + 10\) and \(y = -x + 4\).A* Challenge
Q15Find the set of values of \(k\) for which \(x^2 + (k - 2)x + (k + 1) = 0\) has two distinct real roots.A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \((3, 5)\) and \((-2, 0)\)
1
Substitute: \(x^2 - 4 = x + 2 \implies x^2 - x - 6 = 0 \implies (x - 3)(x + 2) = 0\)
2
\(x = 3\) or \(x = -2\). Using \(y = x + 2\): \(x = 3\) gives \(y = 5\); \(x = -2\) gives \(y = 0\)
\((3, 5)\) and \((-2, 0)\)
Q2 — \((0, -1)\) and \((2, 5)\)
1
Substitute: \(x^2 + x - 1 = 3x - 1 \implies x^2 - 2x = 0 \implies x(x - 2) = 0\)
2
\(x = 0\) or \(x = 2\). Using \(y = 3x - 1\): \(x = 0\) gives \(y = -1\); \(x = 2\) gives \(y = 5\)
\((0, -1)\) and \((2, 5)\)
Q3 — \(x < -3\) or \(x > 3\)
1
Factorise: \((x - 3)(x + 3) > 0\). Critical values \(x = 3\), \(x = -3\)
2
Upward parabola, \(> 0\) outside the roots
\(x < -3\) or \(x > 3\)
Q4 — \(-4 \leq x \leq 4\)
1
Factorise: \((x - 4)(x + 4) \leq 0\). Critical values \(x = 4\), \(x = -4\)
2
Upward parabola, \(\leq 0\) between (and including) the roots
\(-4 \leq x \leq 4\)
Q5 — \(x < 0\) or \(x > 4\)
1
Factorise: \(x(x - 4) > 0\). Critical values \(x = 0\), \(x = 4\)
2
Upward parabola, \(> 0\) outside the roots
\(x < 0\) or \(x > 4\)
Exam-Style (Q6–Q10)
Q6 — \((0, -1)\) and \((4, 3)\)
1
Substitute: \(x^2 - 3x - 1 = x - 1 \implies x^2 - 4x = 0 \implies x(x - 4) = 0\)
2
\(x = 0\) or \(x = 4\). Using \(y = x - 1\): \(x = 0\) gives \(y = -1\); \(x = 4\) gives \(y = 3\)
\((0, -1)\) and \((4, 3)\)
Q7 — \(-\dfrac{3}{2} < x < 5\)
1
\(ac = -30\); numbers \(-10\), \(+3\): \(2x^2 - 10x + 3x - 15 < 0\)
2
Factorise by grouping: \((2x + 3)(x - 5) < 0\). Critical values \(x = -\dfrac{3}{2}\), \(x = 5\)
3
Upward parabola, \(< 0\) between the roots
\(-\dfrac{3}{2} < x < 5\)
Q8 — \(k < -4\) or \(k > 4\)
1
Two distinct real roots: discriminant \(> 0\): \(k^2 - 4(1)(4) > 0\)
2
\(k^2 - 16 > 0\), so \(k^2 > 16\)
\(k < -4\) or \(k > 4\)
Q9 — \(-1 < x < 1\)
1
Factorise: \(x^2 + 2x - 3 = (x + 3)(x - 1) < 0\), so \(-3 < x < 1\)
2
Combine with \(x > -1\): the overlap of \(-3 < x < 1\) and \(x > -1\) is \(-1 < x < 1\)
\(-1 < x < 1\)
Q10 — \(c = 1\)
1
Substitute: \(x^2 - 3x + 5 = x + c \implies x^2 - 4x + (5 - c) = 0\)
2
Tangent means discriminant = 0: \(16 - 4(5 - c) = 0\)
3
\(16 - 20 + 4c = 0\), so \(4c = 4\)
\(c = 1\)
A* Challenge (Q11–Q15)
Q11 — \(-\dfrac{1}{2} < x < 1\), or \(x > 4\)
1
\(x^2 - 5x + 4 = (x - 1)(x - 4) > 0\), so \(x < 1\) or \(x > 4\)
2
\(2x + 1 > 0\), so \(x > -\dfrac{1}{2}\)
3
Combine: \((x < 1\) and \(x > -\tfrac{1}{2})\) gives \(-\dfrac{1}{2} < x < 1\); \((x > 4\) and \(x > -\tfrac{1}{2})\) gives \(x > 4\)
\(-\dfrac{1}{2} < x < 1\), or \(x > 4\)
Q12 — \(k > \dfrac{3}{4}\)
1
Substitute: \(x^2 - x + 3 = 2x + k \implies x^2 - 3x + (3 - k) = 0\)
2
Two distinct points: discriminant \(> 0\): \(9 - 4(3 - k) > 0\)
3
\(9 - 12 + 4k > 0\), so \(4k > 3\)
\(k > \dfrac{3}{4}\)
Q13 — \(\{x: -2 \leq x \leq 4\}\)
1
Factorise: \((x - 4)(x + 2) \leq 0\). Critical values \(x = 4\), \(x = -2\)
2
Upward parabola, \(\leq 0\) between (and including) the roots
\(\{x: -2 \leq x \leq 4\}\)
Q14 — \((2, 2)\) and \((3, 1)\)
1
Substitute: \(x^2 - 6x + 10 = -x + 4 \implies x^2 - 5x + 6 = 0 \implies (x - 2)(x - 3) = 0\)
2
\(x = 2\) or \(x = 3\). Using \(y = -x + 4\): \(x = 2\) gives \(y = 2\); \(x = 3\) gives \(y = 1\)
\((2, 2)\) and \((3, 1)\)
Q15 — \(k < 0\) or \(k > 8\)
1
Discriminant: \((k - 2)^2 - 4(1)(k + 1) > 0\)
2
Expand: \(k^2 - 4k + 4 - 4k - 4 > 0\)
3
Simplify: \(k^2 - 8k > 0\)
4
Factorise: \(k(k - 8) > 0\), an upward parabola in \(k\), positive outside its roots
\(k < 0\) or \(k > 8\)
Summary
- Solve a linear-quadratic system by substituting the linear equation into the quadratic one, then use the linear equation to find each y-value.
- The discriminant of the resulting quadratic tells you the number of intersection points: two distinct roots (two points), repeated root (tangent), no real roots (no intersection).
- Solve quadratic inequalities by rearranging to one side, factorising for critical values, then using the parabola's shape — never by dividing through by the unknown.
- An upward parabola is positive outside its roots and negative between them; a downward parabola is the reverse.
- Combined inequalities need the overlap (intersection) of both regions, not everything from either one.
These techniques build directly on Quadratics — the discriminant reasoning here is the same tool used there, applied to a new type of problem, and feed forward into Graphs & Transformations, where points of intersection are found the same way.
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