How Do Complex Numbers Arise as Solutions to a Quadratic Equation With a Negative Discriminant?

A quadratic equation with a negative discriminant has no real solutions because it requires the square root of a negative number. By defining the imaginary unit as $i=\sqrt{-1}$, these solutions can be expressed as complex numbers.

The Mechanism

For a quadratic equation

$ax^2+bx+c=0,$

the quadratic formula gives

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

The expression $\Delta=b^2-4ac$ is the discriminant. It determines the nature of the roots.

If $\Delta<0$, write $\Delta=-k$, where $k>0$. Then

$\sqrt{\Delta}=\sqrt{-k}=\sqrt{-1}\sqrt{k}=i\sqrt{k}.$

Therefore, the solutions become

$x=\frac{-b\pm i\sqrt{k}}{2a}.$

For example, solve $x^2+4x+13=0$:

$\Delta=4^2-4(1)(13)=16-52=-36.$

Substituting into the quadratic formula,

$x=\frac{-4\pm\sqrt{-36}}{2}=\frac{-4\pm6i}{2}=-2\pm3i.$

The roots are $-2+3i$ and $-2-3i$. They form a complex conjugate pair because the original quadratic has real coefficients: the real parts are equal, while the imaginary parts have opposite signs.

Why the Roots Come in Pairs

For a quadratic with real coefficients, non-real roots must occur as conjugates. If $p+qi$ is a root, substituting its conjugate (p-qi) changes the sign of every term containing $i$, while all real terms remain unchanged. Therefore, the conjugate also satisfies the same equation.

This also matches the graph. A quadratic with $\Delta<0$ has no $x$-intercepts, so its graph never crosses the real $x$-axis. The complex roots still solve the algebraic equation, but they cannot be shown as intercepts on an ordinary Cartesian graph. They can instead be represented as points on an Argand diagram: $-2+3i$ and $-2$ are reflections in the real axis.

A common misconception is that a negative discriminant means the quadratic has “no solutions.” It has no real solutions, but it has two solutions in the complex number system.

IB Exam Technique

This is HL-only content under AHL 1.12, Introduction to complex numbers, applicable to GDC-active Mathematics AI papers. Show the discriminant, use $i^2=-1$, simplify fully into the form $a+bi$, and include both roots from the $\pm$ sign.