Use the discriminant to classify roots and parameter values
Connect the square-root term in the quadratic formula with real intersections, including repeated roots.
Pure Mathematics 1 (Paper 1) · AS & A Level Mathematics 9709, 2026–2027. Paper 1 is part of every route. Written practice is self-checked; it does not award official examination marks.
Before you begin
- Identify signed coefficients in ax² + bx + c = 0.
- Solve a linear inequality, reversing its sign when dividing by a negative number.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What are a, b and c for 2x² − 3x + 5 = 0?
What you will learn
- Use b² − 4ac to count distinct real roots for a ≠ 0.
- Translate a root or intersection condition into a condition on a parameter.
Count roots before calculating them
For real a, b and c with a ≠ 0, the quadratic formula is x = (−b ± √(b² − 4ac))/(2a). Its discriminant D = b² − 4ac determines the real possibilities. D > 0 gives two distinct real roots; D = 0 gives one repeated real root x = −b/(2a); D < 0 gives no real roots. 'No real roots' does not mean the polynomial is undefined, nor does it mean there are no complex roots.
Graphically, a quadratic y = ax² + bx + c crosses the x-axis twice, touches it at the vertex, or misses it. A repeated root is counted twice algebraically but corresponds to one x-coordinate. The rule assumes a non-zero quadratic coefficient: if a depends on a parameter, check values making a = 0 separately because the equation may then be linear.
Apply the rule to the equation you actually solved
For x² − 6x + k = 0, D = 36 − 4k. Requiring two distinct real roots means 36 − 4k > 0, hence k < 9. Equality gives k = 9 and repeated root x = 3; k > 9 gives no real roots. The phrase 'real roots' without 'distinct' includes the repeated case and uses D ≥ 0.
For a line meeting a quadratic curve, equate the two y-expressions and rearrange to a quadratic equal to zero. Use the discriminant of this new equation, not the curve's original x-intercept equation. 'Touches' corresponds to D = 0 for the resulting quadratic; 'two intersections' corresponds to D > 0.
Pause and explain
Which condition gives two distinct real roots of a genuine quadratic?
Worked example
For x² − 6x + k = 0, find when there are two distinct real roots, a repeated real root, or no real roots.
Show the worked solution
- Here a = 1, b = −6, c = k, so D = (−6)² − 4k = 36 − 4k.
- Solve D > 0 to get k < 9, and D = 0 to get k = 9.
- For k > 9, D < 0; at k = 9 the equation is (x − 3)² = 0.
Answer Two distinct real roots: k < 9. Repeated root x = 3: k = 9. No real roots: k > 9.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Classify the real roots of 3x² − 4x + 2 = 0 without solving it.
Give me a hint
Use b = −4 and c = +2 in b² − 4ac.
Compare my reasoning
- Identify a = 3, b = −4, c = 2.
- Calculate D = (−4)² − 4(3)(2) = 16 − 24 = −8.
- A negative discriminant gives no real roots; the graph does not meet the x-axis.
No real roots, because D = −8.
Look for these in your work
- Keep the squared negative coefficient positive.
- State the type of roots from the sign of D.
How many distinct real roots does 2x² + 3x − 2 = 0 have?
Give me a hint
The negative constant makes the product 4ac negative.
Compare my reasoning
- Identify a = 2, b = 3, c = −2.
- Calculate D = 9 − 4(2)(−2) = 25.
- Since 25 > 0, the equation has two distinct real roots.
Two distinct real roots; D = 25.
Look for these in your work
- Subtract the negative product correctly.
- Do not confuse a square discriminant with a repeated root.
The line y = x + k meets y = x² − 3x + 5. Find the values of k giving two intersections, one tangency or no intersections.
Give me a hint
Equate the y-values before calculating a discriminant.
Compare my reasoning
- Rearrange x + k = x² − 3x + 5 to x² − 4x + (5 − k) = 0.
- Its discriminant is 16 − 4(5 − k) = 4k − 4.
- Positive, zero and negative discriminants correspond respectively to k > 1, k = 1 and k < 1. At k = 1, the point is (2, 3).
Two intersections for k > 1; tangent at (2, 3) for k = 1; none for k < 1.
Look for these in your work
- Use the intersection equation's coefficients.
- Translate all three discriminant cases into parameter conditions.
Common mistakes
- Square the signed coefficient using brackets: (−6)² = 36.
- Distinguish real roots from distinct real roots when choosing ≥ or >.
- Check whether a parameter can make the leading coefficient zero.
What can you explain now?
For which real k does x² + 2kx + 9 = 0 have a repeated root?
Compare with the explanation
k = −3 or k = 3.
The discriminant is 4k² − 36. Setting it to zero gives k² = 9, so include both signs.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain the difference between a repeated root and two distinct roots. Then say which equation's discriminant you need when a line meets a parabola.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.