This algebra 2 quadratics complex solutions lesson explains quadratic equations with complex solutions in one printable page, with definitions, a method and a worked example.
A quadratic equation with real coefficients has complex solutions when its discriminant b² − 4ac is negative. The quadratic formula then gives two conjugate solutions p ± qi, because the square root of a negative number is i times a real square root. The graph never touches the x-axis.
Algebra 2 Quadratics Complex Solutions Lesson: key facts
- The discriminant is D = b² − 4ac.
- D > 0: two real solutions; D = 0: one repeated real solution; D < 0: two complex solutions.
- For D < 0, √D = i√(−D), so the solutions have the form p ± qi.
- Complex solutions of a real quadratic always come as a conjugate pair.
- Square roots, completing the square, and the formula all reach the same solutions.
What does the discriminant tell you?
For \(ax^{2}+bx+c=0\) with real coefficients and \(a\neq 0\), the discriminant is \(D=b^{2}-4ac\). Its sign decides the kind of solutions.
| Discriminant | Solutions | Graph |
|---|---|---|
| \(D > 0\) | two different real numbers | crosses the x-axis twice |
| \(D=0\) | one repeated real number | touches the x-axis once |
| \(D < 0\) | two complex numbers, conjugates of each other | never meets the x-axis |
Computing \(D\) first costs one line and tells you what to expect. For \(x^{2}-6x+5=0\), \(D=16\) is positive, so the solutions are real. For \(x^{2}-6x+13=0\), \(D=-16\) is negative, so the solutions are complex. The graph of the related function shows the same fact: it either crosses, touches, or misses the x-axis.
How do you use the quadratic formula when D is negative?
\(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). When \(D < 0\), write \(\sqrt{D}=i\sqrt{-D}\). The two solutions are the complex conjugates \(p+qi\) and \(p-qi\).
Worked example: solve \(x^{2}-4x+13=0\).
- Identify \(a=1\), \(b=-4\), \(c=13\).
- Compute \(D=(-4)^{2}-4(1)(13)=16-52=-36\). It is negative, so expect complex solutions.
- Substitute: \(x=\frac{4\pm\sqrt{-36}}{2}=\frac{4\pm 6i}{2}\).
- Divide both parts by 2: \(x=2\pm 3i\).
When can you solve with square roots or completing the square?
If the equation has the form \(x^{2}=-k\) with \(k > 0\), then \(x=\pm i\sqrt{k}\). For \((x+3)^{2}=-4\) take the square root of both sides to get \(x+3=\pm 2i\), so \(x=-3\pm 2i\). To solve \(x^{2}+6x+13=0\), move the constant, add \(9\) to both sides to get \((x+3)^{2}=-4\), and finish the same way.
Why do the solutions come in pairs?
Always check a complex solution by substituting it and using \(i^{2}=-1\). For \(2+3i\) in \(x^{2}-4x+13\), \((2+3i)^{2}=-5+12i\), and \((-5+12i)-4(2+3i)+13=0\). The formula gives \(p\pm qi\), so complex solutions of a real-coefficient quadratic always appear as a conjugate pair. In the complex plane they are mirror images across the real axis. For \(x^{2}+2x+5=0\), \(D=-16\) and \(x=-1\pm 2i\).
Quadratic Equations with Complex Solutions: frequently asked questions
What does a negative discriminant mean?
It means the quadratic has no real solutions and exactly two complex solutions. The parabola does not cross the x-axis. For \(x^{2}+2x+5=0\) the discriminant is \(-16\).
Why are complex solutions always a conjugate pair?
The quadratic formula has the shape \(\frac{-b}{2a}\pm\frac{\sqrt{D}}{2a}\). When \(D < 0\) the second part is imaginary, and the plus and minus versions are conjugates.
Can I check a complex solution?
Yes. Substitute it and use \(i^{2}=-1\). For \(2+3i\) in \(x^{2}-4x+13\): \((-5+12i)-8-12i+13=0\).
What if the discriminant is zero?
Then \(\sqrt{D}=0\) and the formula gives one solution, \(x=-\frac{b}{2a}\). It is repeated, and the vertex of the parabola sits on the x-axis.
Is the discriminant useful before solving?
Yes. It tells you what to expect, so you can check your answer. A negative D means a result such as \(3\pm 4i\), not real numbers.
How to use this algebra 2 quadratics complex solutions lesson
Use this algebra 2 quadratics complex solutions lesson before the worksheet, then check what you learned with the quiz.
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