The discriminant is the bit under the square root in the quadratic formula — written as Δ = b2 − 4ac. Its sign tells you instantly whether a quadratic has two roots, one root, or no real roots. It’s also the key to questions about unknown coefficients.
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What you need to know
The discriminant is Δ = b2 − 4ac — it’s in the formula booklet
Δ > 0 → two distinct real roots (graph crosses x-axis twice)
Δ = 0 → one repeated real root (graph is tangent to x-axis)
Δ < 0 → no real roots (graph never touches x-axis)
“Real roots” without specifying number → use Δ ≥ 0
Watch for the words distinct vs just real — they give different inequalities
What is the Discriminant?
For a quadratic ax2 + bx + c, the discriminant is:
Δ = b2 − 4ac
It’s the part hiding under the square root in the quadratic formula:
x = −b ± √Δ2a
The sign of Δ decides whether the square root gives two values, one value, or nothing real.
The Three Cases
The sign of Δ tells you everything about the roots:
Δ > 0
Positive
2 distinct real roots
Graph crosses the x-axis at two different points.
Δ = 0
Zero
1 repeated real root
Graph touches the x-axis once — it’s a tangent.
Δ < 0
Negative
No real roots
Graph sits wholly above (or wholly below) the x-axis.
Why this works: the quadratic formula has √Δ. If Δ > 0, you get a real positive square root → two solutions. If Δ = 0, the ± part adds nothing → one solution. If Δ < 0, you can’t take the square root of a negative real number → no real solutions.
Decoding the Question
Exam questions almost never use the word “discriminant” — you have to recognise the situation. Here’s how to translate common phrases:
Question Phrase → Discriminant Condition
“Two distinct real roots” / “crosses x-axis twice”
→
Δ > 0
“Repeated root” / “tangent to x-axis” / “one real solution”
→
Δ = 0
“No real roots” / “never crosses x-axis”
→
Δ < 0
“Real roots” (no number specified)
→
Δ ≥ 0
Watch the wording: “two real roots” can mean Δ ≥ 0 (which includes the repeated case), but “two real distinct roots” strictly means Δ > 0. One word changes the answer.
Finding Unknown Coefficients
The classic exam question gives you a quadratic with an unknown constant (usually called k) and tells you something about its roots. Your job is to find k.
Identify a, b, c in terms of k.
Write down the discriminantΔ = b2 − 4ac and simplify.
Translate the condition on the roots into Δ > 0, Δ = 0, Δ < 0, or Δ ≥ 0.
Solve the resulting equation or inequality for k.
Worked Examples
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Example 1 — Classify the roots
Without solving, state the number of real roots of 2x2 + 3x + 5 = 0.
Answer:
Step 1: identify a, b, c.a = 2, b = 3, c = 5Step 2: compute Δ.Δ = 3² − 4(2)(5) = 9 − 40 = −31Step 3: classify.Δ < 0 → no real roots.No real rootsThe graph of y = 2x² + 3x + 5 sits entirely above the x-axis.
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Example 2 — Show a quadratic has a repeated root
Show that x2 − 6x + 9 = 0 has a repeated real root, and find it.
Answer:
Step 1: identify a, b, c.a = 1, b = −6, c = 9Step 2: compute Δ.Δ = (−6)² − 4(1)(9) = 36 − 36 = 0Step 3: Δ = 0 → repeated root.Step 4: find the root using x = −b/(2a).x = −(−6)/(2 × 1) = 3Repeated root at x = 3
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Example 3 — Find k for repeated roots
Find the values of k for which x2 + kx + 9 = 0 has a repeated real root.
Answer:
Step 1: a = 1, b = k, c = 9.Step 2: write down Δ.Δ = k² − 4(1)(9) = k² − 36Step 3: repeated root → Δ = 0.k² − 36 = 0k² = 36k = 6 or k = −6Both values give a quadratic that touches the x-axis at exactly one point.
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Example 4 — Find k for no real roots
Find the set of values of k for which x2 + 4x + k = 0 has no real roots.
Answer:
Step 1: a = 1, b = 4, c = k.Step 2: write down Δ.Δ = 4² − 4(1)(k) = 16 − 4kStep 3: no real roots → Δ < 0.16 − 4k < 0−4k < −16k > 4 (sign FLIPS when dividing by −4)k > 4Don’t forget to flip the inequality when dividing by a negative.
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Example 5 — Two distinct real roots (full problem)
A function is given by f(x) = 2kx2 + kx − k + 2, where k is a constant. The graph of y = f(x) has two distinct real roots.
(a) Show that 9k2 − 16k > 0.
Step 1: identify a, b, c.a = 2k, b = k, c = −k + 2Step 2: write down Δ = b² − 4ac.Δ = k² − 4(2k)(−k + 2)= k² − 8k(−k + 2)= k² + 8k² − 16k= 9k² − 16kStep 3: two distinct real roots → Δ > 0.9k² − 16k > 0 ✓
(b) Hence find the set of possible values of k.
Step 1: solve 9k² − 16k = 0 to find boundary roots.k(9k − 16) = 0k = 0 or k = 16/9Step 2: sketch the ∪-shape with roots at 0 and 16/9.Step 3: we want > 0 (above x-axis), so pick OUTSIDE the roots.k < 0 or k > 16/9Note: k = 0 would make the original function linear, not quadratic — so it’s excluded automatically.
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Tips
Watch carefully for “distinct”. “Two distinct real roots” needs Δ > 0. “Two real roots” alone could mean Δ ≥ 0. One word makes a different mark.
Always start by writing down a, b, c — including their signs. This step alone catches half the mistakes.
Simplify Δ as much as possible before applying the condition. The answer becomes much cleaner.
For inequalities in k, follow the 4-step quadratic-inequality method: roots, sketch, region.
Don’t forget to flip the inequality sign when multiplying or dividing by a negative number.
Sketch a quick parabola showing where it crosses the x-axis — this confirms which case you’re in visually.
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Common mistakes
Confusing “real roots” with “distinct real roots”. Strict (> 0) vs inclusive (≥ 0) — different inequality, different answer.
Sign errors in b2. If b = −5, then b2 = +25 (a square is always positive, even of a negative).
Forgetting the −4ac sign. If ac is negative, then −4ac is positive — and adds to Δ. Easy to mess up.
Treating the discriminant condition as an equation when it should be an inequality (or vice versa).
Forgetting that Δ > 0 in k is itself a quadratic inequality. Solve it the same way as any other quadratic inequality.
Not flipping the sign when dividing by a negative number while solving for k. Sneaky and very common.
Final word: The discriminant turns root-counting into a one-line calculation. Memorise the three cases, watch for the word “distinct”, and always sketch when solving for unknown k.
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