# Math Help - Radical Equations #3

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?

2. ## 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 $\sqrt{j+\frac{1}{3}}=\sqrt\frac{-1}{3}$

3. ## Re: Radical Equations #3

yes,

I wrote it down correctly, can you comment on my attempt?

4. ## Re: Radical Equations #3

Originally Posted by Mathnood768
solve
My attempt:

Make it $3j+1=12j^2+12j+3$. It is easier to work with.

5. ## Re: Radical Equations #3

Hello, Mathnood768!

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

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