Question

Find the zeros of: f(x) = 2x^2 + 2x + 7

Answer 1

set the function equal to zero and solve for x

Answer 2

The answer my teacher gave: \[\frac{-1\pm \sqrt{13}i}{2}\] though i don't think it's right. If it is, can you show me the steps to achieve this?

Answer 3

2x^2 + 2x + 7 = 0
use quadratic formula to solve for x

Answer 4

Is the answer I provided correct?

Answer 5

That seems about right to me, but I think it could have gone further.

Answer 6

I typed it into mathway.com, and got this http://www.mathway.com/answer.aspx?p=prec?p=f(x)SMB15SMB01SMB152xSMB072SMB15+SMB152xSMB15+SMB157?p=148?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=FindSMB15theSMB15RootsSMB10ZerosSMB15ofSMB15theSMB15Function

Answer 7

ur answer is correct

Answer 8

I just don't understand how to get that answer.

Answer 9

I know to set a,b,c. then solve it out, but I think i'm doing something wrong.

Answer 10

lol. Sorry, I always factor out the 1/2
But, give me one second I'll show ya!

Answer 11

Thanks! I'll gladly wait :)

Answer 12

\[ax^2+bx+c\]
quadratic formula: \[x= \frac{ -b+-\sqrt{(b)^2-4(a)(c)} }{ 2a }\]
Using the quadratic formula, solve for x.
\[x= \frac{ -2+-\sqrt{(2)^2-4(2)(7)} }{ 2*2 }\]
\[x= \frac{ -2+-\sqrt{(2)-56} }{ 4 }\]
\[x= \frac{ -2+-\sqrt{-52} }{ 4 }\]
Now, you have to express
\[\sqrt{-52}\]
in terms of i
\[\sqrt{-52}\]=
\[\sqrt{-1}\sqrt{52}\]
=\[i \sqrt{52}\]
\[x= \frac{ -2+-i\sqrt{52} }{ 4 }\]
simplify radicals
\[\sqrt{52}=\sqrt{4*13}=\sqrt{2^2*13}=2\sqrt{13}\]
\[x= \frac{ -2+-2i\sqrt{13} }{ 4 }\]
and the +- is (plus or minus)

Answer 13

There's more...

Answer 14

You could factor 2 from \[-2+2i \sqrt{13}\]
giving \[2(i \sqrt{13}-1)\]
and then doing the same thing on the other one
so, \[2(-i \sqrt{13}-1)\]
Now, at this time you divide by the 4, and you'll get
\[x=\frac{ -(i \sqrt{13)}-1 }{ 2 } or x=\frac{ i \sqrt{13}-1 }{ 2 }\]

Answer 15

You could also factor out 1/2, but that'd be pointless.

Answer 16

AND You could have also used completing the square method, which is eight times faster and easier.

Answer 17

holy crap thank you! I UNDERSTAND!