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5.10: Solving Quadratic Equations with Complex Number Solutions

Difficulty Level: At Grade / At Grade / Advanced Created by: CK-12
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Miss Harback writes the equation 5x^2 + 125 = 0 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

x^2-4 &= 0\\x &= -2,2

Double root

x^2+4x+4 &= 0\\x &= -2,-2

2 imaginary solutions

x^2+4 &= 0\\x &= -2i,2i

Example A

Solve 3x^2+27=0.

Solution: First, factor out the GCF.

3(x^2+9)=0

Now, try to factor x^2 + 9. Rewrite the quadratic as x^2 + 0x + 9 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.

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

Quadratic equations with imaginary solutions are never factorable.

Example B

Solve (x-8)^2=-25

Solution: Solve using square roots.

(x-8)^2 &= -25\\x-8 &= \pm 5i\\x &= 8 \pm 5i

Example C

Solve 2(3x-5)+10=-30.

Solution: Solve using square roots.

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

Intro Problem Revisit To solve 5x^2+125=0, we first need to factor out the GCF.

5(x^2+25)=0

Now, try to factor x^2 + 25. Rewrite the quadratic as x^2 + 0x + 25 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.

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

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

Guided Practice

1. Solve 4(x-5)^2+49=0.

2. Solve -\frac{1}{2}(3x+8)^2-16=2.

Answers

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

1. 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

2. -\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

Practice

Solve the following quadratic equations.

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

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Difficulty Level:

At Grade

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Date Created:

Mar 12, 2013

Last Modified:

Feb 23, 2014
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