Complex Roots of Quadratics (DP IB Applications & Interpretation (AI): HL): Revision Note

Complex roots of quadratics

When does a quadratic have complex roots?

• The quadratic equation $a{z}^2+bz+c=0$ where $a,b,c\in \mathrm{ℝ}$ has complex roots if

  • $b^2-4ac<0$

    • the discriminant is negative

How do I solve a quadratic with complex roots?

• To solve a quadratic equation with complex roots, $a{z}^2+bz+c=0$

  • either use the quadratic formula

    • e.g. $\pm \sqrt{-9}=\pm 3\mathrm{i}$

  • or complete the square

How are complex roots related to complex conjugates?

• The two complex roots to the quadratic equation $a{z}^2+bz+c=0$ where $a,b,c\in \mathrm{ℝ}$ are complex conjugate pairs

  • i.e. if $z=p+q\mathrm{i}$ is a root, then $z^{*}=p-q\mathrm{i}$ is the other root

• This is not true if any of the coefficients $a$, $b$ or $c$ are complex

How do I factorise a quadratic expression using complex roots?

• If a quadratic expression $a{z}^2+bz+c$ has a negative discriminant ($b^2-4ac<0$), then it can factorised as follows:

  • set the expression equal to zero

    • $a{z}^2+bz+c=0$

  • solve this equation to find the complex roots

    • $z_1=p+q\mathrm{i}$ and $z_2={z_1}^{*}=p-q\mathrm{i}$

  • rewrite $a{z}^2+bz+c$ in the factorised form $a\left(z-z_1\right)\left(z-z_2\right)$

    • You could expand inside each bracket

    • $a\left(z-p-q\mathrm{i}\right)\left(z-p+q\mathrm{i}\right)$