# Quadratic Equations with Complex Solutions

1. Solving Equations
2. Discriminant Use
3. Examples

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

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

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

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.

NOTE:

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.

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 SolutionsExamples

(1.1)

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

Solution

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}}}.

(1.2)

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

Solution

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}}}

(1.3)

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

Solution

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}}

1. Home
2.  ›
3. Algebra 2