is it possible to solve: 7d^2+12=0??

Question

saifoo.khan · Answer

Yes!

saifoo.khan · Answer

but the solutions will not be real.

saifoo.khan · Answer

$$d^2 = -\frac{12}{7} $$Now take square root on both sides.

anonymous · Answer

If we assume that $d$ is a real number, then no.  If it is allowed to be an imaginary or complex number, then most certainly!

anonymous · Answer

okay. i must've done the previous steps wrong. because the answer is -2 and 2

anonymous · Answer

umm can you solve 10d-2t=-40? then substitute it into 7d^2+5d-t-8=0 pleasse?

saifoo.khan · Answer

d and t?

anonymous · Answer

yeah

anonymous · Answer

WolframAlpha is a good resource to check your answers: http://www.wolframalpha.com/input/?i=10d-2t%3D-40+and+7d%5E2%2B5d-t-8%3D0

anonymous · Answer

Check your work carefully... your troubles were likely caused by an erroneous slip of the pen.  (It happens to all of us!)

anonymous · Answer

thanks, but it doesn't show how to solve it algebraically. which I need to do

jim_thompson5910 · Answer

10d-2t=-40

-2t=-40-10d

t = -40/(-2) - 10d/(-2)

t = 20+5d <<--- You probably made the error here

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7d^2+5d-t-8=0

7d^2+5d-(20+5d)-8=0

7d^2+5d-20-5d-8=0

7d^2-28=0


Now solve for d

anonymous · Answer

ohhh ok. yeah i didnt know how to solve for d then I figured it out...thanks!

jim_thompson5910 · Answer

you're welcome