Hidden under the square root of the quadratic formula is one number, \(b^{2} - 4ac\). Its sign tells you whether a quadratic has two roots, one repeated root or none, and it turns questions about unknown constants into inequalities.
For \(ax^{2} + bx + c = 0\): if \(b^{2} - 4ac > 0\) there are two distinct real roots; if \(b^{2} - 4ac = 0\) there is one repeated root; if \(b^{2} - 4ac < 0\) there are no real roots. 'Real roots' (not necessarily distinct) means \(b^{2} - 4ac \ge 0\).
Fig. 1: three parabolas with the same shape, where Δ = b² - 4ac. The curve with Δ > 0 crosses the x-axis twice, Δ = 0 touches it once, Δ < 0 misses it
Section 01
One number, three outcomes.
The quadratic formula \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) only gives real answers when the number under the root is not negative. That number is the discriminant. Before you calculate it, rearrange the equation to \(ax^{2} + bx + c = 0\) and read \(a\), \(b\) and \(c\) WITH their signs.
\(b^{2} - 4ac\)
Roots of \(ax^{2} + bx + c = 0\)
Graph of \(y = ax^{2} + bx + c\)
\(> 0\)
two distinct real roots
crosses the \(x\)-axis twice
\(= 0\)
one repeated (equal) root
touches the \(x\)-axis at its vertex
\(< 0\)
no real roots
does not meet the \(x\)-axis
Worked example Classify three equations
How many real roots does each equation have? (a) \(3x^{2} - 5x + 1 = 0\) (b) \(4x^{2} - 12x + 9 = 0\) (c) \(2x^{2} + 3x + 4 = 0\)
(a) \(a = 3,\ b = -5,\ c = 1\): \(b^{2} - 4ac = 25 - 12 = 13 > 0\), so two distinct real roots.
(b) \(a = 4,\ b = -12,\ c = 9\): \(144 - 144 = 0\), so one repeated root.
Here \(4x^{2} - 12x + 9 = (2x - 3)^{2}\), so the repeated root is \(x = \dfrac{3}{2}\).
(c) \(a = 2,\ b = 3,\ c = 4\): \(9 - 32 = -23 < 0\), so no real roots.
Answer: (a) two (b) one repeated root, \(x = \frac{3}{2}\) (c) none
Worked example Watch the order of terms
Find the discriminant of \(5 - x - 2x^{2}\).
Reorder: \(-2x^{2} - x + 5\), so \(a = -2,\ b = -1,\ c = 5\).
\(b^{2} - 4ac = (-1)^{2} - 4(-2)(5) = 1 + 40\).
Answer: \(41\), so \(5 - x - 2x^{2} = 0\) has two distinct real roots
Examiner trap
\(b^{2}\) is never negative: with \(b = -5\), \(b^{2} = 25\), not \(-25\). Bracket negative values when you substitute: \((-5)^{2} - 4(3)(1)\).
Section 02
Finding an unknown constant.
Most exam questions give a quadratic containing a constant \(k\) and describe its roots. Translate the words into a condition on the discriminant, then solve for \(k\).
Equal / repeated root (or "the curve touches the \(x\)-axis"): \(b^{2} - 4ac = 0\).
Two distinct real roots: \(b^{2} - 4ac > 0\).
Real roots: \(b^{2} - 4ac \ge 0\).
No real roots: \(b^{2} - 4ac < 0\).
Worked example Equal roots
Find the values of \(k\) for which \(kx^{2} + 4x + (k - 3) = 0\) has equal roots, and find the root for each value.
\(a = k,\ b = 4,\ c = k - 3\). Equal roots: \(16 - 4k(k - 3) = 0\).
Divide by \(-4\): \(k^{2} - 3k - 4 = 0\), so \((k - 4)(k + 1) = 0\).
Read the wording exactly. 'Real roots' includes equal roots, so it needs \(\ge 0\); 'two distinct real roots' needs strict \(> 0\). Using the wrong one costs the final mark.
Stage 1 · Learn
Check your understanding
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What is the discriminant of \(2x^{2} - 3x - 5\)?
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If \(b^{2} - 4ac = 0\), the equation \(ax^{2} + bx + c = 0\) has
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How many times does \(y = x^{2} + 4x + 5\) meet the \(x\)-axis?
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\(x^{2} + 6x + k = 0\) has a repeated root. Find \(k\).
Section 03
Always positive, always real.
The discriminant also proves facts that hold for every \(x\) or every value of a constant. A quadratic with \(a > 0\) and \(b^{2} - 4ac < 0\) never touches the \(x\)-axis, so its graph lies entirely above it (Fig. 1, top curve).
Worked example Positive for all x
Find the set of values of \(k\) for which \(kx^{2} + 6x + k > 0\) for all real \(x\).
The parabola must open upwards, so \(k > 0\).
It must not meet the \(x\)-axis: \(36 - 4k \cdot k < 0\), so \(k^{2} > 9\).
\(k^{2} > 9\) means \(k > 3\) or \(k < -3\); combined with \(k > 0\) only \(k > 3\) remains.
Answer: \(k > 3\)
Worked example Real roots for every p
Show that \(x^{2} + px + p - 1 = 0\) has real roots for all values of \(p\).
A square is never negative, so \(b^{2} - 4ac \ge 0\) for every \(p\).
Answer: The roots are always real (and equal only when \(p = 2\))
Examiner trap
For 'positive for all \(x\)' the condition \(b^{2} - 4ac < 0\) is not enough on its own: \(-x^{2} - 1\) has a negative discriminant but is always NEGATIVE. You also need \(a > 0\).
Stage 2 · Exam
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Find the values of \(k\) for which the equation \(x^{2} + kx + 2k - 3 = 0\) has equal roots.
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Find the set of values of \(k\) for which \(3x^{2} - 6x + k = 0\) has two distinct real roots.
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Find the set of values of \(m\) for which \(2x^{2} + mx + 8 = 0\) has no real roots.
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Which statement about \(x^{2} + (k + 1)x + k = 0\) is true?
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