Complex Numbers and Quadratic Equations • Topic 3 of 3

Quadratic Equations with Complex Roots

A quadratic equation has the form $ax^2 + bx + c = 0$ with $a \ne 0$. Its roots are given by the quadratic formula, and the quantity under the square root — the discriminant $D = b^2 - 4ac$ — decides the nature of those roots.

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

Over the real numbers, a negative discriminant means "no solution". Once complex numbers exist, that gap closes: a negative $D$ simply produces a square root of a negative number, which we write using $i$. The result is a pair of complex roots in the form $a + ib$.

Discriminant $D = b^2 - 4ac$Nature of roots
$D > 0$Two distinct real roots
$D = 0$One repeated real root
$D < 0$Two complex conjugate roots

When $D < 0$, write $\sqrt{D} = \sqrt{-(4ac - b^2)} = i\sqrt{4ac - b^2}$, so the roots become:

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

For an equation with real coefficients, complex roots always arrive in conjugate pairs: if $p + iq$ is a root, then $p - iq$ is the other. This is why a quadratic over the reals can never have exactly one non-real root.

Relations between roots and coefficients still hold, and they hold even when the roots are complex:

$$\alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}$$

These let you check an answer quickly: for conjugate roots $p \pm iq$, the sum is the real number $2p$ and the product is $p^2 + q^2$, both real, exactly as $-b/a$ and $c/a$ must be.

Deeper Insight — why complex roots are a feature, not a failure: A negative discriminant used to mark the end of the road, but it is really a signpost that the answers live one dimension over, off the real line. The reason this works so cleanly is the conjugate pairing forced by real coefficients: the imaginary parts of $p + iq$ and $p - iq$ cancel in the sum and combine into the real quantity $p^2 + q^2$ in the product, so the coefficients $-b/a$ and $c/a$ stay perfectly real even though the roots are not. This is a small instance of the Fundamental Theorem of Algebra, which promises that a degree-$n$ polynomial has exactly $n$ roots in the complex numbers, counted with multiplicity — no equation is ever truly unsolvable. The practical lesson is to stop treating $D < 0$ as an error and start reading it as an instruction: factor out the $i$, simplify the surd, and report the conjugate pair.

The discriminant decides the nature of the roots The Discriminant Decides D > 02 real roots D = 01 repeated root D < 0no real root2 complex roots Complex roots occur as conjugate pairs on the Argand plane Complex Roots Come in Conjugate Pairs ReIm p + iq p − iq mirror images across the real axis
1
Worked Example
Solve $x^2 + 1 = 0$.
Solution
  1. Rearrange: $x^2 = -1$.
  2. Take square roots: $x = \pm\sqrt{-1} = \pm i$.

Answer: $x = i$ or $x = -i$

2
Worked Example
Solve $x^2 + 4 = 0$.
Solution
  1. Rearrange: $x^2 = -4$.
  2. $x = \pm\sqrt{-4} = \pm\sqrt{4}\,i = \pm 2i$.

Answer: $x = 2i$ or $x = -2i$

3
Worked Example
Solve $x^2 - 2x + 5 = 0$.
Solution
  1. Here $a = 1,\ b = -2,\ c = 5$, so $D = (-2)^2 - 4(1)(5) = 4 - 20 = -16$.
  2. $\sqrt{D} = \sqrt{-16} = 4i$.
  3. $x = \dfrac{-(-2) \pm 4i}{2(1)} = \dfrac{2 \pm 4i}{2}$.
  4. Simplify: $x = 1 \pm 2i$.

Answer: $x = 1 + 2i$ or $x = 1 - 2i$

4
Worked Example
Solve $2x^2 + x + 1 = 0$.
Solution
  1. $a = 2,\ b = 1,\ c = 1$, so $D = 1^2 - 4(2)(1) = 1 - 8 = -7$.
  2. $\sqrt{D} = \sqrt{-7} = i\sqrt{7}$.
  3. $x = \dfrac{-1 \pm i\sqrt{7}}{2(2)} = \dfrac{-1 \pm i\sqrt{7}}{4}$.

Answer: $x = -\dfrac{1}{4} \pm \dfrac{\sqrt{7}}{4}i$

5
Worked Example
Solve $x^2 - 4x + 13 = 0$ and verify the sum and product of its roots.
Solution
  1. $a = 1,\ b = -4,\ c = 13$, so $D = 16 - 52 = -36$ and $\sqrt{D} = 6i$.
  2. $x = \dfrac{4 \pm 6i}{2} = 2 \pm 3i$.
  3. Sum of roots: $(2 + 3i) + (2 - 3i) = 4 = -\dfrac{b}{a} = -\dfrac{-4}{1}$. ✓
  4. Product: $(2 + 3i)(2 - 3i) = 4 - 9i^2 = 4 + 9 = 13 = \dfrac{c}{a}$. ✓

Answer: $x = 2 \pm 3i$; sum $= 4$, product $= 13$, both verified.

6
Worked Example
Solve $\sqrt{3}\,x^2 - \sqrt{2}\,x + 3\sqrt{3} = 0$.
Solution
  1. $a = \sqrt{3},\ b = -\sqrt{2},\ c = 3\sqrt{3}$.
  2. $D = (-\sqrt{2})^2 - 4(\sqrt{3})(3\sqrt{3}) = 2 - 36 = -34$.
  3. $\sqrt{D} = i\sqrt{34}$, and $2a = 2\sqrt{3}$.
  4. $x = \dfrac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}$.

Answer: $x = \dfrac{\sqrt{2} \pm i\sqrt{34}}{2\sqrt{3}}$

Key Points

  • For $ax^2 + bx + c = 0$, the discriminant $D = b^2 - 4ac$ fixes the nature of the roots.
  • $D > 0$ gives two real roots, $D = 0$ a repeated real root, and $D < 0$ a pair of complex roots.
  • When $D < 0$, write $\sqrt{D} = i\sqrt{|D|}$ and read off the roots in $a + ib$ form.
  • With real coefficients, non-real roots always occur as conjugate pairs $p \pm iq$.
  • Check using $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$ — both stay real for conjugate roots.
Tap an option to check your answer0 / 4
Q1.The roots of $x^2+1=0$ are:
Explanation: $x^2=-1\Rightarrow x=\pm i$.
Q2.If the discriminant $b^2-4ac<0$, the roots are:
Explanation: Negative discriminant gives complex conjugate roots.
Q3.The roots of $x^2-2x+2=0$ are:
Explanation: $x=\tfrac{2\pm\sqrt{-4}}{2}=1\pm i$.
Q4.The sum of the roots of $ax^2+bx+c=0$ is:
Explanation: Sum of roots $=-b/a$.