Question

what is the solution to 2x^2+x+2=0 ?

Answer 1

Did you try use quadratic formula?

Answer 2

i did but my final answer was not any of my answer choices :/ ?

Answer 3

Maybe simplify?

What are choices and what is your answer?

Answer 4

I would use the completing the square method.

Answer 5

I got C, how about you?

Answer 6

a=2 b=1 c=2

\[\frac{ -1 \pm \sqrt{1-16} }{ 4 }\]

Answer 7

you get -15 which i square root 15

Answer 8

mine answer are \[x = \frac{ i(\sqrt{15}+ i) }{ 4 }, x = - \frac{ i(\sqrt{15} - i) }{ 4 }\]

Answer 9

negative fraction. everything else looks the same, so i got c.

Answer 10

what happened to the -1?

Answer 11

for -b of the quadratic formula?

Answer 12

oh ive got just a bad headache it with this problem its alright guys ill just skip it thanks for helping out :D

Answer 13

first group your terms.\[(2x^2+x)+2=0\]Factor out a 2.\[2\left(x^2+\frac{x}{2}\right)+2=0\]Fidn your new c value. \(c=\left(\frac{b}{2}\right)^2 = \left(\frac{\frac{1}{2}}{2}\right)^2 =\frac{1}{16}\)\[2\left(x^2 +\frac{x}{2}+\frac{1}{16}\right) +2 -\frac{1}{8}=0\]\[2\left(x+\frac{1}{4}\right)^2+\frac{15}{8}=0 \]

Answer 14

\[2\left(x+\frac{1}{4}\right)^2 = -\frac{15}{8}\]\[\left(x+\frac{1}{4}\right)^2 = -\frac{15}{16}\]\[x+\frac{1}{4} = \frac{i\sqrt{15}}{4}\]\[x=-\frac{1}{4}+\frac{i\sqrt{15}}{4}=\frac{-1+i\sqrt{15}}{4} \]

Answer 15

There is your answer.

Answer 16

Oh, -.- forgot the \(\pm\) didnt I. D'oh.....

Answer 17

\[x= \frac{-1\pm i\sqrt{15}}{4}\]

Answer 18

thank you o.o n.n !