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December 12th, 2015, 08:47 AM   #1
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About square roots in quadratic equations where bx=0

Why is 16^1/2=4, but when x²=16, then x=+/-4?
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December 12th, 2015, 09:11 AM   #2
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$\displaystyle \sqrt{x^2} =|x|$

$\displaystyle x^2=16\Longleftrightarrow \sqrt{x^2} = \sqrt{16}\Longleftrightarrow |x|=4 \Longleftrightarrow x= \pm4 $
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December 12th, 2015, 09:21 AM   #3
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Oh, now I see, thank you!
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December 12th, 2015, 10:20 AM   #4
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Quote:
Originally Posted by renatomoraesp View Post
Why is 16^1/2=4, $\displaystyle \ \ \ \ \ \ $It's not.

but when x²=16, then x=+/-4?
16^1/2 = $\displaystyle \ \dfrac{16^1}{2} \ = \ \dfrac{16}{2} \ = \ 8$



You need grouping symbols:

16^(1/2) = $\displaystyle \sqrt{16} \ = \ 4$
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January 24th, 2016, 01:31 AM   #5
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Oh, thank you, I did not know!
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