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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

Texas Algebra II (TEKS) • Standard 2A.4(H) • Complex Numbers

Complex Solutions to Quadratics is a topic in Complex Numbers in the Texas Essential Knowledge and Skills (Algebra II, §111.40). It is aligned to Standard 2A.4(H), which requires students to solve quadratic equations having complex solutions.

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

Texas Algebra II (TEKS) › Complex Numbers › Complex Solutions to Quadratics  —  Standard 2A.4(H)

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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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