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Quadratic Equations with Complex Solutions

On this Page:
 1. Solving Equations
 2. Discriminant Use
 3. Examples

Complex numbers can be solutions to equations in Math at times.

This page will focus on quadratic equations with complex solutions as an introduction to the topic.

Let’s consider  2  standard quadratics and look at their respective graphs.

y = x^2 \space {\text{–}} \space 1      and      y = x^2 + 1.

Quadratic Equation with real solutions.
Example of when you can have quadratic equations with complex solutions.

It can be seen that the graph of   y = x^2 \space {\text{–}} \space 1   does touch the x-axis and has roots at  2  points.

Where as the graph of   y = x^2 + 1   does not touch the x-axis at any point, it has no real roots.

Solving the Quadratics as Equations

We can write these quadratics as equations equal to  0.

Then solve and find any roots.

      x^2 \space {\text{–}} \space 1 = 0     =>     x^2 = 1     =>     \sqrt{x^2} = \pm \sqrt{1}     =>     x = \pm 1

      x^2 + 1 = 0     =>     x^2 = {\text{-}}1     =>     \sqrt{x^2} = \pm \sqrt{{\text{-}}1}     =>     x = \pm i

As expected for   x^2 \space {\text{–}} \space 1 = 0,  we get the real solutions of  -1  and  1.

But for   x^2 + 1 = 0,  even though there are no real solutions, we do get two complex solutions of  –i  and  i.


Finding two complex solutions together is actually a general case for quadratic equations with complex solutions.
They occur in ‘conjugate pairs’, one with a positive sign and one with a negative sign, which we will see in examples further on in the page.

Making use of the Discriminant:

If we recall the discriminant of a quadratic, this is a value that tells us what the nature of roots of a quadratic equation will be.

The formula for the discriminant is the portion inside the radical of the quadratic formula.

Quadratic Complex Solutions 3

So to recap, for a quadratic equation   ax^2 + bx + c = 0.

The value of the discriminant tells us.

\boldsymbol{{b^{2}} – 4ac}  >  0     There are two different real roots.
\boldsymbol{{b^{2}} – 4ac}  =  0     There are two equal real roots, so one root in total.
\boldsymbol{{b^{2}} – 4ac}  <  0     There are no real roots, instead there are two complex solutions.

So the working discriminant is a handy way of finding out in advance what kind of roots/solutions you can expect to see for a quadratic equation.

Quadratic Equations with Complex Solutions


Find the roots of   x^2 \space {\text{–}} \space 3x + 9 = 0.


a = 1 \space\space , \space\space b = {\text{-}}3 \space\space , \space\space c = 9

We can solve using the quadratic formula.

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}b \space \pm \space \sqrt{b^{2} \space – \space 4ac}}{2a}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{3 \space \pm \space \sqrt{({\text{-}}3^{2} \space\space – \space\space (4 \times 1 \times 9)}}{2 \times 1}}}    =    {\boldsymbol{\frac{3 \space \pm \space \sqrt{9 \space\space {\text{–}} \space\space 36}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{3 \space \pm \space \sqrt{{\text{-}}25}}{2}}}    =    {\boldsymbol{\frac{3 \space \pm \space \sqrt{25 \space \times \space {\text{-}}1}}{2}}}    =    {\boldsymbol{\frac{3 \space \pm \space \sqrt{25}\sqrt{{\text{-}}1}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{3 \space \pm \space 5i}{2}}}
Which we can alternatively write as   {\boldsymbol{\frac{3}{2} {\scriptsize{\pm}} \frac{5i}{2}}}.


Find the roots of   x^2 \space {\text{–}} \space x + 2 = 0.


a = 1 \space\space , \space\space b = {\text{-}}1 \space\space , \space\space c = 2

{\scriptsize{x = \space}} {\boldsymbol{\frac{1 \space \pm \space \sqrt{({\text{-}}1^{2} \space\space – \space\space (4 \times 1 \times 2)}}{2 \times 1}}}    =    {\boldsymbol{\frac{1 \space \pm \space \sqrt{1 \space\space {\text{–}} \space\space 8}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{1 \space \pm \space \sqrt{{\text{-}}7}}{2}}}    =    {\boldsymbol{\frac{1 \space \pm \space \sqrt{7 \space \times \space {\text{-}}1}}{2}}}    =    {\boldsymbol{\frac{1 \space \pm \space \sqrt{7}\sqrt{{\text{-}}1}}{2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{1 \space \pm \space \sqrt{7}i}{2}}}     or     {\boldsymbol{\frac{1}{2} {\scriptsize{\pm}} \frac{\sqrt{7}i}{2}}}


Find the roots of   {\text{-}}x^2 + 4x \space {\text{–}} \space 6 = 0.


a = {\text{-}}1 \space\space , \space\space b = 4 \space\space , \space\space c = {\text{-}}6

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{({\text{-}}4)^{2} \space\space – \space\space (4 \times ({\text{-}}1) \times ({\text{-}}6))}}{2 \space \times \space ({\text{-}}1)}}}    =    {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{16 \space\space {\text{–}} \space\space 24}}{{\text{-}}2}}}

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{{\text{-}}8}}{{\text{-}}2}}}    =    {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{8 \space \times \space {\text{-}}1}}{{\text{-}}2}}}    =    {\boldsymbol{\frac{{\text{-}}4 \space \pm \space \sqrt{8}\sqrt{{\text{-}}1}}{{\text{-}}2}}}

( \bf{\sqrt{8}}  can be simplified further to  \bf{2\sqrt{2}}. )

{\scriptsize{x = \space}} {\boldsymbol{\frac{{\text{-}}4 \space \pm \space 2\sqrt{2}i}{{\text{-}}2}}}    =    {\boldsymbol{\frac{{\text{-}}4}{{\text{-}}2} \space {\scriptsize{\pm}} \space \frac{2\sqrt{2}i}{{\text{-}}2}}}

x = \space {\boldsymbol{2 \space \pm \sqrt{2}i}}

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