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This question appeared in my homework, and we went over it in class today.

Solve the Zeros:

$$3x^2 = -81$$

My solution: $$3x^2=-81\\3x=\pm9i\\x=\pm3i$$

Correct Solution: $$3x^2=-81\\x^2=-27\\x=\sqrt{-27}\\x=\sqrt{9}\sqrt{3}\sqrt{-1}\\x=\pm3i\sqrt{3}$$

Why is the text book's solution correct, and not mine?

Community
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Travis
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    You failed to take the square root of $3$. – lulu Jan 10 '17 at 15:41
  • When you took the square root of both sides, you left 3 untouched. Do you see how the book divided both sides by 3 before taking the square root of both sides? Try getting rid of constants before taking square roots. – scott Jan 13 '17 at 16:22

2 Answers2

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In your solution, when you extracted the square root of on both sides of your equation, you forgot to extract the square root of 3 on the left side. $$\sqrt{3x^2} \neq 3x$$

The correct extraction would be:

$$\sqrt{3x^2} = \sqrt3x$$

This was your only mistake.

From now on, try getting rid of constants before extracting a square root, if possible. Do this until you're familiar with this type of operations.

theSongbird
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$$3x^2 = -81 \implies \sqrt 3x = \pm 9 i$$

Argon
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