Miss Harback writes the equation \begin{align*}5x^2 + 125 = 0\end{align*}

Corrine says the equation has two real solutions. Drushel says the equation has a double root, so only one solution. Farrah says the equation has two imaginary solutions.

Which one of them is correct?

### Quadratic Equations with Complex Number Solutions

When you solve a quadratic equation, there will always be two answers. Until now, we thought the answers were always real numbers. In actuality, there are quadratic equations that have imaginary solutions as well. The possible solutions for a quadratic are:

2 real solutions

\begin{align*}x^2-4 &= 0\\
x &= -2,2\end{align*}

Double root

\begin{align*}x^2+4x+4 &= 0\\
x &= -2,-2\end{align*}

2 imaginary solutions

\begin{align*}x^2+4 &= 0\\
x &= -2i,2i\end{align*}

#### Solve the following problems

Solve \begin{align*}3x^2+27=0\end{align*}

First, factor out the GCF.

\begin{align*}3(x^2+9)=0\end{align*}

Now, try to factor \begin{align*}x^2 + 9\end{align*}

\begin{align*}3x^2+27 &= 0\\
3x^2 &= -27\\
x^2 &= -9\\
x &= \pm \sqrt{-9}= \pm 3i\end{align*}

*Quadratic equations with imaginary solutions are never factorable.*

Solve \begin{align*}(x-8)^2=-25\end{align*}

Solve using square roots.

\begin{align*}(x-8)^2 &= -25\\
x-8 &= \pm 5i\\
x &= 8 \pm 5i\end{align*}

Solve \begin{align*}2(3x-5)+10=-30\end{align*}

Solve using square roots.

\begin{align*}2(3x-5)^2+10 &=-30\\
2(3x-5)^2 &=-40\\
(3x-5)^2 &=-20\\
3x-5 &=\pm2i\sqrt{5}\\
3x &=5\pm2i\sqrt{5}\\
x &=\frac{5}{3}\pm\frac{2\sqrt{5}}{3}i\end{align*}

### Examples

#### Example 1

Earlier, you were asked which one of them is correct.

To solve \begin{align*}5x^2+125=0\end{align*}

\begin{align*}5(x^2+25)=0\end{align*}

Now, try to factor \begin{align*}x^2 + 25\end{align*}

\begin{align*}5x^2 + 125 &= 0\\
5x^2 &= -125\\
x^2 &= -25\\
x &= \pm \sqrt{-5}= \pm 5i\end{align*}

The equation has two roots and both of them are imaginary, so Farrah is correct.

#### Example 2

Solve \begin{align*}4(x-5)^2+49=0\end{align*}

\begin{align*}4(x-5)^2+49 &=0\\
4(x-5)^2 &=-49\\
(x-5)^2 &=-\frac{49}{4}\\
x-5 &=\pm\frac{7}{2}i\\
x &=5\pm\frac{7}{2}i\end{align*}

#### Example 3

Solve \begin{align*}-\frac{1}{2}(3x+8)^2-16=2\end{align*}

\begin{align*}-\frac{1}{2}(3x+8)^2-16 &=2\\
-\frac{1}{2}(3x+8)^2 &=18\\
(3x+8)^2 &=-36\\
3x+8 &=\pm6i\\
3x &=-8\pm6i\\
x &=-\frac{8}{3}\pm2i\end{align*}

Both of these quadratic equations can be solved by using square roots.

### Review

Solve the following quadratic equations.

- \begin{align*}x^2=-9\end{align*}
x2=−9 - \begin{align*}x^2+8=3\end{align*}
x2+8=3 - \begin{align*}(x+1)^2=-121\end{align*}
(x+1)2=−121 - \begin{align*}5x^2+16=-29\end{align*}
- \begin{align*}14-4x^2=38\end{align*}
- \begin{align*}(x-9)^2-2=-82\end{align*}
- \begin{align*}-3(x+6)^2+1=37\end{align*}
- \begin{align*}4(x-5)^2-3=-59\end{align*}
- \begin{align*}(2x-1)^2+5=-23\end{align*}
- \begin{align*}-(6x+5)^2=72\end{align*}
- \begin{align*}7(4x-3)^2-15=-68\end{align*}
- If a quadratic equation has \begin{align*}4 - i\end{align*} as a solution, what must the other solution be?
- If a quadratic equation has \begin{align*}6 + 2i\end{align*} as a solution, what must the other solution be?
**Challenge**Recall that the factor of a quadratic equation has the form \begin{align*}(x\pm m)\end{align*} where \begin{align*}m\end{align*} is any number. Find a quadratic equation that has the solution \begin{align*}3 + 2i\end{align*}.- Find a quadratic equation that has the solution \begin{align*}1-i\end{align*}.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 5.10.