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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! Tags bx0, equations, quadratic, roots, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post haitensionBB Algebra 6 July 24th, 2015 04:01 PM watkd Algebra 9 August 5th, 2012 09:59 PM fantom.1040 Algebra 2 July 2nd, 2011 06:04 AM watkd Algebra 5 August 25th, 2010 10:43 AM romeroom Algebra 7 March 17th, 2009 01:53 PM

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