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 April 19th, 2018, 02:44 AM #1 Newbie   Joined: Mar 2018 From: Canada Posts: 14 Thanks: 0 equation with exponent Hi everyone, I just have a doubt when solving the following. Here is what I've done. 2x^2=18 x^2=18/2 x^2=9 x=sqrt9 x=3 My doubt is if the answer is only 3 or -3 is also an answer because of the square root. Thank you
 April 19th, 2018, 03:02 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 It's easy to check, isn't it? 3^2= 9, 2(9)= 18 so 3 is a solution. (-3)^2= 9, 2(9)= 18 so -3 is also a solution. Both 3 and -3 are solutions.
April 19th, 2018, 03:04 AM   #3
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 Originally Posted by Country Boy It's easy to check, isn't it? 3^2= 9, 2(9)= 18 so 3 is a solution. (-3)^2= 9, 2(9)= 18 so -3 is also a solution. Both 3 and -3 are solutions.
thank you! I just wanted to make sure !

 April 19th, 2018, 03:48 AM #4 Senior Member   Joined: Oct 2009 Posts: 770 Thanks: 276 Note though that $\sqrt{9} = 3$ and not $-3$. The square root is by definition always positive. However, the solutions to $x^2=9$ are both $3$ and $-3$. Crucial distinction! Thanks from Country Boy
April 19th, 2018, 08:07 AM   #5
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 Originally Posted by Micrm@ss Note though that $\sqrt{9} = 3$ and not $-3$. The square root is by definition always positive. However, the solutions to $x^2=9$ are both $3$ and $-3$. Crucial distinction!
thanks. That actually clarifies a lot!

 April 20th, 2018, 11:09 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 That's why you need the "$\displaystyle \pm$" when you say that "the solution to $\displaystyle x^2= a$ is $\displaystyle x= \pm\sqrt{a}$".

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