# Radical Equations #3

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• Jun 20th 2013, 10:58 PM
Mathnood768
Radical Equations #3
http://puu.sh/3kF3d.pngsolve

My attempt:

http://puu.sh/3kF5o.png
http://puu.sh/3kF6T.png
http://puu.sh/3kFaS.png

when I put it into the quadratic formula: http://puu.sh/3kFcp.png

the answer is suppose to be j = -2/3 what did i do wrong?
• Jun 21st 2013, 12:08 AM
downthesun01
Re: Radical Equations #3
Are you sure you copied this down right? I'm getting a negative radical when I use the quadratic equation.

Right away, plugging in $\frac{-2}{3}$ makes $\sqrt{j+\frac{1}{3}}=\sqrt\frac{-1}{3}$
• Jun 21st 2013, 12:26 PM
Mathnood768
Re: Radical Equations #3
yes, http://puu.sh/3kF3d.png

I wrote it down correctly, can you comment on my attempt?
• Jun 21st 2013, 12:52 PM
Plato
Re: Radical Equations #3
Quote:

Originally Posted by Mathnood768

Make it $3j+1=12j^2+12j+3$. It is easier to work with.
• Jun 21st 2013, 03:27 PM
Soroban
Re: Radical Equations #3
Hello, Mathnood768!

Quote:

I wrote it down correctly. . No, I don't think so.

If the answer is $j = \text{-}\tfrac{2}{3}$
. . the equation could be: . $\sqrt{\frac{j+1}{3}} + 5j \:=\: 3i - 1$