Question

Which of the following is a solution of x2 + 2x + 8?

I know it is supposed to contain an Imaginary number, but I am having trouble getting through it. Can someone help me? Thank you!

Answer 1

Do you know the quadratic formula?

Answer 2

Yes

Answer 3

Perhaps you are meant to use another method, but quadratic formula is easiest.

Answer 4

Alright, I understand the quadratic formula. How would I set it up?
Thanks!

Answer 5

\[
x^2 + 2x + 8 = (1)x^2+(2)x+(8)
\]

Answer 6

\[
a=1,b=2,c=8
\]

Answer 7

\[
x = \frac{-b\pm \sqrt{b^2-4ac}}{2a} = \frac{-(2)\pm \sqrt{(2)^2-4(a)(8)}}{2(1)}
\]

Answer 8

I ended with \[-2\pm \sqrt{-14}\]

Answer 9

did you divide by 2 ? (2a)

Answer 10

That is correct, but you can simplify it.

Answer 11

Oh!
I see!

Answer 12

Oh, yeah, wait, divide by 2.

Answer 13

\[-1\pm \sqrt{-7}\]?

Answer 14

not quite

Answer 15

how and what did you get for inside the square root ? can you show your steps?

Answer 16

Well, I multiplied 2^2 and got 4.
Then I multiplied 4(1)(8) and got 32.
I subtracted 4-32.

Answer 17

I got -28 and divided that by 2 to get -14.
@phi

Answer 18

\[
\frac{\sqrt{-28}}{2}=\frac{\sqrt{-28}}{\sqrt{4}} = \sqrt{-28/4} =\sqrt{-7}
\]

Answer 19

ok with \( \sqrt{-28} \)
but you don't divide the "stuff" inside by 2
however, you can write -28= -4*7
so sqr(-28) = sqr(4*-7) = sqr(4) * sqr(-7)

Answer 20

or do as wio showed: to "move" the 1/2 inside the square root, square it to get 1/4
and move it inside to get sqr(-28/4)

Answer 21

Alright, so to do what you did Wio, I just move it inside the square root and square it. Getting a 4 rather than a 2.
Thanks so much, both of you. I think I really understand this better. It took 20 minutes, thank you so much for your time! I can't thank you enough!

Answer 22

so you should get
\[ \frac{-2 \pm \sqrt{-28}}{2} \\
= -1 \pm \sqrt{-7} \\
= -1 \pm \sqrt{7}\ i
\]

Answer 23

Yes! I did! Thank you both!