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Algebra 2 Complex numbers

Complex solutions to quadratics

20 practice questions 0 video lessons Theory + worked examples

Complex Solutions to Quadratics

California Algebra 2 • Standard N-CN.7 • Complex Numbers

Complex Solutions to Quadratics is a topic in Complex Numbers in the California Common Core State Standards. It is aligned to Standard N-CN.7, which requires students to solve quadratic equations with real coefficients that have complex solutions.

When the discriminant is negative, a quadratic has two complex conjugate roots found with the quadratic formula and \(i\).

California Algebra 2 › Complex Numbers › Complex Solutions to Quadratics  —  Standard N-CN.7

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Theory

When \(D=b^2-4ac<0\), a quadratic has two complex roots. The quadratic formula still applies:

\[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},\]

and a negative radicand becomes \(i\sqrt{|D|}\).

Real quadratics have complex roots in conjugate pairs \(a\pm bi\).
A quadratic with complex roots When a parabola misses the x-axis, its roots are complex conjugates. x y x²-2x+5 no x-intercepts
A parabola with no \(x\)-intercepts has complex roots.
Complex roots Complex roots Complex roots D < 0 → two complex roots quadratic formula still works roots are conjugates a ± bi
Complex roots from a negative discriminant.

The quadratic formula (complex case):

\[x=\dfrac{-b\pm i\sqrt{4ac-b^2}}{2a}\]
the quadratic formula with a negative discriminant gives complex conjugate roots
Simplify \(\sqrt{-k}\) to \(i\sqrt{k}\) before dividing.

How to find complex roots

  1. Write \(ax^2+bx+c=0\) and compute \(D\).
  2. If \(D<0\), apply the quadratic formula.
  3. Rewrite \(\sqrt{D}\) as \(i\sqrt{|D|}\).
  4. Simplify to \(a\pm bi\).
Example 1 — Pure imaginary roots
Solve \(x^2+9=0\).
Solution

Isolate \(x^2\) and take roots.

\(x^2\)\(=\)\(-9\)
\(x\)\(=\)\(\pm3i\)
x equals plus or minus 3 i
Example 2 — Quadratic formula
Solve \(x^2-2x+5=0\).
Solution

Use the quadratic formula with \(D=4-20=-16\).

\(x\)\(=\)\(\dfrac{2\pm\sqrt{-16}}{2}\)
\(=\)\(\dfrac{2\pm4i}{2}\)
\(=\)\(1\pm2i\)
x equals 1 plus or minus 2 i
Example 3 — Another complex pair
Solve \(x^2+4x+13=0\).
Solution

Here \(D=16-52=-36\).

\(x\)\(=\)\(\dfrac{-4\pm\sqrt{-36}}{2}\)
\(=\)\(\dfrac{-4\pm6i}{2}\)
\(=\)\(-2\pm3i\)
x equals negative 2 plus or minus 3 i
Example 4 — Conjugate pairs
If \(2+3i\) is a root of a real quadratic, what is the other root?
Solution

Complex roots of a real polynomial come in conjugate pairs.

\(\text{other root}\)\(=\)\(2-3i\)
the other root is 2 minus 3 i

Common pitfalls

A negative discriminant is not “no solution” — it gives complex roots.
Simplify the radical to \(i\sqrt{k}\) before dividing by \(2a\).
Divide every term by \(2a\), including the real part.

Frequently asked questions

When does a quadratic have complex roots?

When the discriminant \(b^2-4ac\) is negative.

Do complex roots come in pairs?

Yes — for a real quadratic they are conjugates \(a\pm bi\).

Does the quadratic formula still work?

Yes; the negative radicand just introduces \(i\).

How do you simplify \(\sqrt{-16}\)?

\(\sqrt{-16}=4i\).

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