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# 5.10: Solving Quadratic Equations with Complex Number Solutions

Difficulty Level: At Grade Created by: CK-12
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Practice Quadratic Formula and Complex Sums

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Miss Harback writes the equation 5x2+125=0\begin{align*}5x^2 + 125 = 0\end{align*} on the board. She asks the class how many solutions the equation has and what type they are.

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?

### Guidance

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

x24x=0=2,2\begin{align*}x^2-4 &= 0\\ x &= -2,2\end{align*}

Double root

x2+4x+4x=0=2,2\begin{align*}x^2+4x+4 &= 0\\ x &= -2,-2\end{align*}

2 imaginary solutions

x2+4x=0=2i,2i\begin{align*}x^2+4 &= 0\\ x &= -2i,2i\end{align*}

#### Example A

Solve 3x2+27=0\begin{align*}3x^2+27=0\end{align*}.

Solution: First, factor out the GCF.

3(x2+9)=0\begin{align*}3(x^2+9)=0\end{align*}

Now, try to factor x2+9\begin{align*}x^2 + 9\end{align*}. Rewrite the quadratic as x2+0x+9\begin{align*}x^2 + 0x + 9\end{align*} to help. There are no factors of 9 that add up to 0. Therefore, this is not a factorable quadratic. Let’s solve it using square roots.

3x2+273x2x2x=0=27=9=±9=±3i\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.

#### Example B

Solve (x8)2=25\begin{align*}(x-8)^2=-25\end{align*}

Solution: Solve using square roots.

(x8)2x8x=25=±5i=8±5i\begin{align*}(x-8)^2 &= -25\\ x-8 &= \pm 5i\\ x &= 8 \pm 5i\end{align*}

#### Example C

Solve 2(3x5)+10=30\begin{align*}2(3x-5)+10=-30\end{align*}.

Solution: Solve using square roots.

2(3x5)2+102(3x5)2(3x5)23x53xx=30=40=20=±2i5=5±2i5=53±253i\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*}

Intro Problem Revisit To solve 5x2+125=0\begin{align*}5x^2+125=0\end{align*}, we first need to factor out the GCF.

5(x2+25)=0\begin{align*}5(x^2+25)=0\end{align*}

Now, try to factor x2+25\begin{align*}x^2 + 25\end{align*}. Rewrite the quadratic as x2+0x+25\begin{align*}x^2 + 0x + 25\end{align*} to help. There are no factors of 25 that add up to 0. Therefore, this is not a factorable quadratic. Let’s solve it using square roots.

5x2+1255x2x2x=0=125=25=±5=±5i\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.

### Guided Practice

1. Solve 4(x5)2+49=0\begin{align*}4(x-5)^2+49=0\end{align*}.

2. Solve 12(3x+8)216=2\begin{align*}-\frac{1}{2}(3x+8)^2-16=2\end{align*}.

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

1. 4(x5)2+494(x5)2(x5)2x5x=0=49=494=±72i=5±72i\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*}

2. 12(3x+8)21612(3x+8)2(3x+8)23x+83xx=2=18=36=±6i=8±6i=83±2i\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*}

### Practice

Solve the following quadratic equations.

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

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### Vocabulary Language: English

complex number

A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.

complex root

A complex root is a complex number that, when used as an input ($x$) value of a function, results in an output ($y$) value of zero.

Imaginary Numbers

An imaginary number is a number that can be written as the product of a real number and $i$.

The quadratic formula states that for any quadratic equation in the form $ax^2+bx+c=0$, $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$.

Real Number

A real number is a number that can be plotted on a number line. Real numbers include all rational and irrational numbers.

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