Learning Objectives
Classify distinct real roots using the discriminant.
Translate root-count conditions into conditions on a parameter.
Check the leading coefficient before using a quadratic test.
Relate a repeated root to a squared factor.
Prerequisites: Section 1C Lesson 01 quadratic formula and completed squares. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
The discriminant tells you how many distinct real roots a quadratic equation has without requiring you to calculate the roots. This is especially useful when a coefficient depends on a parameter, a fixed value whose possible choices we want to investigate.
We connect the discriminant to the quadratic formula and graph, then turn root-count requirements into conditions on a parameter. Finally, we check exceptional cases where the equation stops being quadratic and explain what a root count cannot tell you.
1. Using the Discriminant to Count Real Roots
The quadratic formula contains a square root. Whether the quantity inside it is positive, zero or negative determines the number of distinct real answers.
1.1 Defining the discriminant and its three cases
For ax² + bx + c = 0 with a ≠ 0, define the discriminant Δ = b² − 4ac. In the quadratic formula, Δ determines the available real square roots. If Δ > 0 there are two distinct real roots, if Δ = 0 there is one distinct repeated root, and if Δ < 0 there are no real roots.
1.2 Distinguishing a repeated root from two distinct roots
“Two real roots” sometimes counts multiplicity, so read the wording carefully. In this lesson, “two distinct roots” excludes the repeated-root case. A repeated root is a value whose factor occurs twice: a(x − r)².
1.3 Connecting root counts to the graph
Graphically, roots are the x-coordinates where y = ax² + bx + c meets the horizontal axis. The diagram shows upward-opening cases; a downward-opening parabola has the same root-count classification.
Figure 1. Two, one repeated, and no real roots. These sketches show root counts rather than numerical coordinates.
Worked Example: Compare Three Equations
Question: Classify x² − 6x + 8 = 0, x² − 6x + 9 = 0 and x² − 6x + 10 = 0.
Method: Calculate the discriminant of each equation and classify its sign, then check the count using completed-square form.
Their discriminants are 4, 0 and −4, so the numbers of distinct real roots are two, one and zero.
Their completed forms are (x − 3)² − 1, (x − 3)² and (x − 3)² + 1, which independently give the same counts.
2. Finding Parameter Values for a Required Root Count
When a coefficient varies, the discriminant varies with it. Translate the requested root count into a sign condition and solve that condition for the parameter.
2.1 Writing a condition on the discriminant
If a coefficient contains a parameter, calculate Δ as an expression in that parameter. Then impose the sign condition required by the question. A boundary where Δ = 0 separates repeated-root behaviour from the neighbouring cases.
2.2 Keeping both branches of a squared-parameter condition
For a squared parameter, a condition such as k² > 16 means k < −4 or k > 4: values with distance from zero greater than 4. Do not keep only the positive branch.
Worked Example: Two Distinct Roots
Question: For which real k does x² + kx + 4 = 0 have two distinct real roots?
Method: Require a strictly positive discriminant and solve the squared-parameter inequality on both sides of zero.
Δ = k² − 16, so we need k² > 16. Therefore k < −4 or k > 4.
At k = ±4 the equation is a perfect square and has a repeated root. At k = 0 it has no real roots, which checks the middle region.
Worked Example: No Real Roots
Question: For which real t does 2x² − 4x + t = 0 have no real roots?
Method: Require a negative discriminant, solve the resulting linear inequality, and check the boundary using a completed square.
Δ = 16 − 8t < 0 gives t > 2.
Completing the square gives 2(x − 1)² + t − 2. If t > 2 it is positive for every real x; at t = 2 it vanishes at x = 1.
3. Checking Whether the Equation Remains Quadratic
Before applying a quadratic rule to every parameter value, check that the squared term is still present. A zero leading coefficient requires a separate calculation.
3.1 Handling the linear exception
The discriminant rule assumes a nonzero quadratic coefficient. If that coefficient can vanish, handle that parameter separately in the original equation. A linear equation can have one solution without having a repeated quadratic root.
3.2 Recognising an identity or contradiction
If all variable terms disappear, the equation can be an identity or a contradiction. Those cases are decided by its remaining constant, not by a quadratic formula with a zero denominator.
Worked Example: A Linear Exception
Question: For which real k does kx² + 2x + 1 = 0 have exactly one distinct real solution?
Method: Test the parameter that removes the quadratic term first. For all other values, impose a zero discriminant.
If k = 0, the equation is 2x + 1 = 0, giving x = −1/2. If k ≠ 0, one distinct real solution requires Δ = 4 − 4k = 0, so k = 1.
At k = 1 the equation is (x + 1)² = 0. The complete parameter set is {0, 1}, with different reasons for the two values.
4. Separating Root Count from Root Location
Knowing that roots exist does not tell you where they lie. Questions about positive roots or a specified interval need additional information.
4.1 Why the discriminant does not determine root signs
A positive discriminant proves two distinct real roots, but does not say whether they are positive or negative. For instance, x² − 5x + 6 and x² + 5x + 6 both have discriminant 1; their roots are 2, 3 and −2, −3 respectively.
4.2 Adding a method that locates the roots
When a problem also asks for roots in a particular interval, use the actual roots, a completed-square form, or a sign argument in addition to the discriminant. Do not infer the interval from Δ alone.
Common Mistakes
Use the full signed coefficient b before squaring it.
Include Δ = 0 only when a repeated root is allowed.
Check a = 0 separately in a parameter problem.
A real-root count does not locate those roots or determine their signs.
5. Practice
Begin each parameter question by checking whether the leading coefficient can be zero. Then use the discriminant condition matching the requested number of distinct real roots, with careful treatment of equality.
1. Classify the real roots of 3x² + 2x + 1 = 0.
2. Find t for which x² − 8x + t = 0 has a repeated root, and find that root.
3. Find k for which x² + 2kx + 9 = 0 has two distinct real roots.
4. Find m for which 5x² + 10x + m = 0 has no real roots.
5. Find all k for which (k − 1)x² + 2x + 1 = 0 has exactly one real solution.
6. Does Δ > 0 imply one positive and one negative root? Give a counterexample.
Worked Solutions
1. Classifying roots from a negative discriminant
Δ = 4 − 12 = −8, so there are no real roots. Equivalently, 3(x + 1/3)² + 2/3 is strictly positive.
2. Finding a repeated-root parameter and root
64 − 4t = 0 gives t = 16. The equation becomes (x − 4)² = 0, so the root is 4.
3. Keeping both parameter intervals
4k² − 36 > 0 gives k² > 9, so k < −3 or k > 3. Equality gives repeated roots, so exclude ±3.
4. Finding a no-real-root condition
100 − 20m < 0 gives m > 5. Completing the square gives 5(x + 1)² + m − 5, which is positive exactly in that parameter range.
5. Combining the linear exception with the repeated-root case
If k = 1, the equation is linear and has root −1/2. Otherwise Δ = 4 − 4(k − 1) = 8 − 4k. Setting it to zero gives k = 2, with equation (x + 1)² = 0. Thus k = 1 or 2.
6. Disproving a claim about root signs
No. x² − 5x + 6 = 0 has Δ = 1 > 0 but both roots, 2 and 3, are positive.
6. Summary
- For a genuine quadratic, the sign of Δ controls the number of distinct real roots.
- Translate the wording into a strict or non-strict condition carefully.
- Parameter values that remove the quadratic term need a separate original-equation check.
- Use more information than the discriminant to locate roots.