solve My attempt: when I put it into the quadratic formula: the answer is suppose to be j = -2/3 what did i do wrong?
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Are you sure you copied this down right? I'm getting a negative radical when I use the quadratic equation. Right away, plugging in $\displaystyle \frac{-2}{3}$ makes $\displaystyle \sqrt{j+\frac{1}{3}}=\sqrt\frac{-1}{3}$
yes, I wrote it down correctly, can you comment on my attempt?
Originally Posted by Mathnood768 solve My attempt: Make it $\displaystyle 3j+1=12j^2+12j+3$. It is easier to work with.
Last edited by Plato; Jun 21st 2013 at 11:55 AM.
Hello, Mathnood768! I wrote it down correctly. . No, I don't think so. If the answer is $\displaystyle j = \text{-}\tfrac{2}{3}$ . . the equation could be: .$\displaystyle \sqrt{\frac{j+1}{3}} + 5j \:=\: 3i - 1$
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