How do I solve this? -3x2 - 4x - 4 = 0 Please help!!
You don't know how to solve quadratic!
there are 3 methods to do this -Complete the square -Quadratic formula -Focatoring
\(\large \rm -3x^2-4x-4=0 ~~\Longrightarrow 3x^2+4x+4=0\) after multiplying out -1 First thing you need to do is test the discriminant \(\large \rm \Delta=b^2-4ac\) if it less than zero then this has complex solution is it positive then it has real solution
I already tried to solve it using the quadratic formula >.< The answer I got was 2/3 (1-sqrtof2) But this is somehow not correct because the only answers that are possible are as follows. And I'm sure what I did wrong :c
*i'm not sure >.<
well that's because you didn't solve it the right way
\(\large \rm \Delta=4^2-4*4*3=16-48=-32\) so we have complex solutions here we need quadratic formula \(\large \rm x=\frac{-4\pm \sqrt{-32}}{6}\) \(\Huge \rm x=\frac{-4\pm \sqrt{(-1)*16*2}}{6}=\frac{-4\pm 4i\sqrt{2}}{6}\) \(\Huge x=\frac{-2\pm 2i\sqrt{2}}{3}\)
@xapproachesinfinity Ahhh! I see what I did wrong... thanks so much!! Umm, do you think you can help me with one more question however, if it isn't too much to ask? ;w;
you are welcome! post and i will see if i can
@xapproachesinfinity Okay! Thanks so much! XD Okay this for this one, since the equation is different than the one from my last question, I wasn't sure if the quadratic formula could still be applied to it >.< But here it is! Solve 3x2 = -12x - 15.
that basically the same thing you have 3x^2=-12x-15 you add 12x to both sides you have 3x^2+12x=-15 then add 15 to both sides you get 3x^2+12x+15=0 then solve
@xapproachesinfinity Ah! I see >.< Am I still supposed to multiply the -1 out? Or is it okay?
OF course not! i did that to get rid of - sign but when you a good expression you don't have to notice that whether i took that sign off or not i will get the same results i just preferred to make it nice
XD Okay!
Okay, is this correct so far? :x
yes good!
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