Quadratic equations and the discriminant

The quadratic formula still works

• For $ax^2 + bx + c = 0$ with $a \neq 0$:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

• Each variable:
  • $a$ — the coefficient of $x^2$
  • $b$ — the coefficient of $x$
  • $c$ — the constant term
• At Level 3 the formula is used even when the square root is negative, because $\sqrt{\text{negative}}$ is now a legal object.

The discriminant

• The discriminant is the expression under the root:

$$\Delta = b^2 - 4ac$$

• It decides the nature of the roots without solving anything:

  • $\Delta > 0$ — two distinct real roots
  • $\Delta = 0$ — one repeated real root (a "double root")
  • $\Delta < 0$ — two distinct complex roots, which are conjugates of each other

• Note the phrase used in exam questions: "real and distinct" means $\Delta > 0$; "equal roots" or "a repeated root" means $\Delta = 0$; "no real roots" means $\Delta < 0$.

• Graphically: $\Delta$ counts how many times the parabola $y = ax^2 + bx + c$ crosses the $x$-axis. When it crosses nowhere, the roots are complex.

Solving when the discriminant is negative

• Apply the quadratic formula as usual.

• Rewrite the negative root using $\sqrt{-n} = i\sqrt{n}$.

• Simplify the surd, then split the fraction into $a + bi$ form.

• Worked through for $x^2 - 4x + 13 = 0$:

  • $\Delta = 16 - 52 = -36$, so the roots are complex
  • $$x = \dfrac{4 \pm \sqrt{-36}}{2} = \dfrac{4 \pm 6i}{2} = 2 \pm 3i$$

• The two roots $2 + 3i$ and $2 - 3i$ are conjugates, as they must be for a quadratic with real coefficients.

Completing the square

• The alternative method, and the one to use when the coefficients contain unknown constants that stop a calculator helping.

• For $x^2 + px + q = 0$:

  • Halve the coefficient of $x$ and square it.
  • Add and subtract that value to build a perfect square.
  • Rearrange and take the square root of both sides — remembering both signs.

• Worked through for $x^2 + 6x + 25 = 0$:

  • $$(x + 3)^2 - 9 + 25 = 0$$
  • $$(x + 3)^2 = -16$$
  • $$x + 3 = \pm 4i$$
  • $$x = -3 \pm 4i$$

Building a quadratic from its roots

• If a quadratic has roots $\alpha$ and $\beta$, then:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

• Sum of roots $= -\dfrac{b}{a}$, product of roots $= \dfrac{c}{a}$.

• Given a complex root of a real-coefficient quadratic, you get the other one free — it is the conjugate — and then:

  • Sum $= (a + bi) + (a - bi) = 2a$, which is real
  • Product $= (a + bi)(a - bi) = a^2 + b^2$, which is real

• So the quadratic always comes out with real coefficients, as it must.

• Worked through: a quadratic with root $3 - 2i$ and real coefficients.

  • Other root: $3 + 2i$
  • Sum: 6. Product: $9 + 4 = 13$.
  • Equation: $x^2 - 6x + 13 = 0$

Questions with an unknown constant

• The 2026 specification states that candidates must handle "the manipulation of constants, e.g. solving an equation to find an expression for $x$ in terms of $k$".
• These questions typically read: "find the values of $k$ for which $x^2 + kx + 9 = 0$ has complex roots".
  • Form the discriminant: $\Delta = k^2 - 36$
  • Apply the condition: complex roots means $\Delta < 0$, so $k^2 - 36 < 0$
  • Solve the inequality: $k^2 < 36$, so $-6 < k < 6$
• Note that the answer is a range, not a single value, because the condition was an inequality.