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 December 12th, 2015, 08:47 AM #1 Newbie     Joined: Feb 2015 From: Brazil Posts: 13 Thanks: 0 Math Focus: Still figuring it out. About square roots in quadratic equations where bx=0 Why is 16^1/2=4, but when x²=16, then x=+/-4?
 December 12th, 2015, 09:11 AM #2 Senior Member     Joined: Apr 2014 From: Europa Posts: 584 Thanks: 177 $\displaystyle \sqrt{x^2} =|x|$ $\displaystyle x^2=16\Longleftrightarrow \sqrt{x^2} = \sqrt{16}\Longleftrightarrow |x|=4 \Longleftrightarrow x= \pm4$
 December 12th, 2015, 09:21 AM #3 Newbie     Joined: Feb 2015 From: Brazil Posts: 13 Thanks: 0 Math Focus: Still figuring it out. Oh, now I see, thank you!
December 12th, 2015, 10:20 AM   #4
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Quote:
 Originally Posted by renatomoraesp 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) = $\displaystyle \sqrt{16} \ = \ 4$

 January 24th, 2016, 01:31 AM #5 Newbie     Joined: Feb 2015 From: Brazil Posts: 13 Thanks: 0 Math Focus: Still figuring it out. Oh, thank you, I did not know!

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