01 · Count roots
How many distinct real roots does 4x² + 3x + 2 = 0 have?
Hint
Calculate b² − 4ac with the coefficient signs intact.
Worked solution
D = 9 − 32 = −23 < 0
No real roots.
Understand · explore · practise
The discriminant
Use b² − 4ac to count real roots, solve parameter questions, explain tangency and prove when roots must exist.
Before you startThe quadratic formula and basic inequalities
01 / Count roots
The square root controls the possibilities.
In the quadratic formula, the part under the square-root sign is the discriminant, usually written D or Δ.
D = b² − 4ac for ax² + bx + c = 0
The leading coefficient a must be non-zero.
- D > 0: two distinct real roots. The graph crosses the x-axis twice.
- D = 0: one repeated real root, x = −b/(2a). The graph touches the x-axis at its vertex.
- D < 0: no real roots. There are no x-axis intersections.
“Distinct” means different. A repeated root counts twice algebraically, but there is only one distinct real solution.
Move c through 4 in y = x² − 4x + c. The curve moves vertically: two crossings meet, then disappear.
y = x² − 4x + cMove c
c = 1. D = 16 − 4c = 12. Two distinct real roots: 2 − √3 and 2 + √3.
Watch two roots meet and disappear
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Calculate
Identify coefficients after rearranging.
Bring the equation to zero before finding a, b and c. A missing term has coefficient zero. Keep each sign attached to its coefficient.
D counts real roots without calculating them. It does not say whether the roots are positive, negative or rational.
Does multiplying the equation change the answer?
Multiplying all coefficients by a non-zero number λ changes the discriminant to λ²D. Its sign is unchanged, so the number of real roots is unchanged.
(λb)² − 4(λa)(λc)
= λ²(b² − 4ac)
Compare three equationsWorked example
3x² + x − 2 = 0
D = 1 + 24 = 25
Two distinct real roots.
9x² − 12x + 4 = 0
D = 144 − 144 = 0
One repeated root: x = 12/18 = 2/3.
−2x² + 4x − 5 = 0
D = 16 − 40 = −24
No real roots. A negative leading coefficient does not reverse the discriminant rules.
03 / Find a parameter
Translate the words into a condition.
Replace “equal roots”, “two distinct real roots” or “no real roots” by D = 0, D > 0 or D < 0 before doing the algebra.
Equal roots
x² + kx + 16 = 0
D = k² − 64
D = 0 ⇒ k = ±8
There are two values of the parameter. Each produces a different perfect-square quadratic.
Two distinct roots
x² − 6x + k = 0
36 − 4k > 0 ⇒ k < 9
At k = 9 the roots merge, so that endpoint is excluded.
“Has real roots” allows a repeated root: use D ≥ 0. If “two roots” is ambiguous, distinguish two distinct roots from roots counted with repetition.
A range with two intervalsWorked example
x² + kx + 9 = 0
Find k for two distinct real roots.
D = k² − 36 > 0
Factor the expression in k.
(k − 6)(k + 6) > 0
Both factors have the same sign outside the two critical values.
k < −6 or k > 6
Do not replace this by k > 6 alone. Negative k of large magnitude also works.
04 / Prove & check
A positive discriminant still needs a quadratic.
To prove a discriminant is always positive, complete the square in the parameter. A non-negative square plus a positive constant cannot be zero or negative.
f(x) = 2x² + (k + 2)x − k
D = (k + 2)² + 8k
= (k + 6)² − 32
This particular discriminant is not always positive: at k = −6 it is −32. The claim must follow from the expression, not from a few favourable examples.
A valid always-positive proof
g(x) = x² + (k + 1)x − k² − 1
D = 5k² + 2k + 5
= 5(k + 1/5)² + 24/5 > 0
The leading coefficient is 1 for every k. Therefore g has two distinct real roots for every real k.
Can b always be chosen to give real roots?
Fix non-zero a and c. Since b² can be made as large as needed, choosing b with b² > 4ac always gives two distinct real roots. Equal roots require b² = 4ac, which has a real b only when ac > 0 under these non-zero assumptions.
The exceptional parameter valueWorked example
kx² + (k + 1)x + 1 = 0
The leading coefficient vanishes when k = 0.
D = (k + 1)² − 4k = (k − 1)²
For k ≠ 0, it is a quadratic.
k = 1: (x + 1)² = 0
One repeated root. For k ≠ 0, 1 there are two distinct real roots.
k = 0: x + 1 = 0
This is linear, with one solution. Do not apply the quadratic formula or root-count rule here.
05 / Tangency
Apply the test to the intersection equation.
A line touches a parabola at exactly one point when the equation for their intersections has a repeated root. Form that equation first, then set its discriminant to zero.
The discriminant of the parabola alone only counts intersections with the x-axis. It does not count intersections with every possible line.
Connect the vertex to the discriminant
ax² + bx + c
= a(x + b/(2a))² − D/(4a)
The vertex height is −D/(4a). If D = 0 it lies on the x-axis. When D < 0, an upward parabola lies wholly above the axis and a downward parabola wholly below it.
Find the tangent in a parallel familyWorked example
Parabola: y = x² + 2x + 3
Line: y = 4x + k
At an intersection, equate the two expressions.
x² − 2x + 3 − k = 0
This is the quadratic to test.
D = 4 − 4(3 − k) = 4k − 8
D = 0 ⇒ k = 2
The tangent line is y = 4x + 2.
x² − 2x + 1 = 0 ⇒ x = 1
y = 6
The contact point is (1, 6).
06 / Your turn
Write the condition before solving.
For parameter questions, check whether the leading coefficient can be zero. State strict or inclusive endpoints explicitly.
02 · A repeated root
Find q so 3x² + 6x + q = 0 has a repeated root; then find that root.
Hint
Set D = 0, then use −b/(2a).
Worked solution
36 − 12q = 0 ⇒ q = 3
x = −6/6 = −1
The equation becomes 3(x + 1)² = 0.
03 · No real roots
Find k so 2x² − 5x + k = 0 has no real roots.
Hint
The discriminant must be negative, not merely non-positive.
Worked solution
25 − 8k < 0
k > 25/8
Equality gives a repeated root and is excluded.
04 · Two parameter intervals
Find p so x² + px + 4 = 0 has real roots, allowing a repeated root.
Hint
Use D ≥ 0 and remember the negative values of p.
Worked solution
p² − 16 ≥ 0
p ≤ −4 or p ≥ 4
At p = ±4 the root is repeated.
05 · Check a claim
A student says tx² + 3x + 1 = 0 has two distinct real roots whenever t < 9/4. Correct the statement.
Hint
D = 9 − 4t, but what happens to the degree at t = 0?
Worked solution
For t < 9/4 and t ≠ 0, it is a quadratic with D > 0 and two distinct real roots. At t = 0 it becomes 3x + 1 = 0, with only x = −1/3.
06 · Find a tangent
Find k for which y = 2x + k is tangent to y = x² − 4x + 7. Find the contact point.
Hint
The intersection equation is x² − 6x + 7 − k = 0.
Worked solution
D = 36 − 4(7 − k) = 8 + 4k
D = 0 ⇒ k = −2
(x − 3)² = 0 ⇒ x = 3
y = 2(3) − 2 = 4
The contact point is (3, 4).
07 / Recap
Count roots before calculating them.
- D = b² − 4ac applies to a genuine quadratic, a ≠ 0.
- Positive: two distinct real roots. Zero: one repeated real root. Negative: no real roots.
- Translate the condition precisely before solving for a parameter.
- Check separately any parameter value that makes the equation linear.
- For a line and a curve, test the equation formed by equating their outputs.
Section 1 of 7 · Count roots