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 September 29th, 2012, 05:54 PM #1 Newbie   Joined: Sep 2012 Posts: 4 Thanks: 0 Help with Algebra Question Have a problems with question Make x the subject: y=2?(3x^2-1 ) I've tried lots of different ways but I can't quite get the answer. I think I have to get rid of the ? first. That's what I'm not sure about. Anyway the answer is: x=?((y^2+ 4)/12 )
 September 29th, 2012, 06:23 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: Help with Algebra Question $y\,=\,2\sqrt{3x^2\,-\,1}$ $y^2\,=\,4(3x^2\,-\,1)$ $\frac{y^2}{4}\,=\,3x^2\,-\,1$ $\frac{y^2}{4}\,=\,3$$x^2\,-\,\frac13$$$ $\frac{y^2}{12}\,=\,x^2\,-\,\frac13$ $\frac{y^2}{12}\,+\,\frac13\,=\,x^2$ $x\,=\,\pm\sqrt{\frac{y^2}{12}\,+\,\frac13}\,=\,\pm \sqrt{\frac{y^2\,+\,4}{12}},\,|x|\,\ge\,\frac{1}{\ sqrt{3}}$ The above accounts for the possibility of negative x.
September 29th, 2012, 06:31 PM   #3
Math Team

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From: Lexington, MA

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Re: Help with Algebra Question

Hello, walker8476!

Quote:
 $\text{Solve for }x:\;\; y \:=\:2\sqrt{3x^2\,-\,1}$ $\text{Answer: }\:x \;=\;\sqrt{\frac{y^2\,+\,4}{12}}$

Take it one step at a time.

Look at what was being done to the x.

(1) x is squared.
(2) Then multiplied by 3.
(3) Then 1 is subtracted.
(4) Then the square root is taken.
(5) Then the result is multiplied by 2.

Now, we read up the list, using the inverse operations.

$\text{W\!e have: }\:2\sqrt{3x^2\,-\,1} \:=\:y$

$\text{(1) Divide by 2:}
\;\;\;\sqrt{3x^2\,-\,1} \:=\:\frac{y}{2}$

$\text{(2) Square:}
\;\;\;\left(\sqrt{3x^2\,-\,1}\right)^2 \;=\;\left(\frac{y}{2}\right)^2 \;\;\;\Rightarrow\;\;\;3x^2\,-\,1 \;=\;\frac{y^2}{4}$

$\text{(3) Add 1:}
\;\;\;3x^2 \;=\;\frac{y^2}{4}\,+\,1 \;\;\;\Rightarrow\;\;\;3x^2 \;=\;\frac{y^2\,+\,4}{4}$

$\text{(4) Divide by 3:}
\;\;\;x^2 \;=\;\frac{y^2\,+\,4}{12}$

$\text{(5) Take the square root:}
\;\;\;x \;=\;\sqrt{\frac{y^2\,+\,4}{12}}$

 September 30th, 2012, 03:09 PM #4 Newbie   Joined: Sep 2012 Posts: 4 Thanks: 0 Re: Help with Algebra Question Thank you both. I understand now. I like the approach of writing out the steps then going up the list. I never knew about this approach. Everything is simple when you break it down!

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