Here's the question you clicked on:
tvise
b^2+16b+1=0 is the answer b=8+3root7?
\[\large{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\] are you sure that you applied this formula?
It says complete the square do i still do the quadratic formula?
If requests that you complete the square, you should do so. I think you have it almost whipped, check your signs.
no if it says complete the square then : \[\large{b^2+16b+1=0}\] \[\large{b^2+16b+1+64=0+64}\] \[\large{(b+8)^2 = 63}\] \[\large{b = \sqrt{63} -8}\]
I came up with:\[-8\pm3\sqrt{7}\]
\[\large{b = -8 \pm \sqrt{63}}\] \[\large{b = -8 \pm 3\sqrt{7}}\]
Note that \[\sqrt{63}=\sqrt{9}\sqrt{7}\]
Yes. When i originally did it i did not isolate the variable so when i re did it i forgot to change signs
no problem, take care of it from now, :)
Good luck, and the "signs" can bite you. lol