Question

g(x) =12x^2 + 12x + 15
How do I find the solutions? I know it will have two real irrational solutions

Answer 1

Find the gcf first.
In this case it is 3.
3(4x^2 + 4x + 5)

Answer 2

Oh, okay. I should have known that haha

Answer 3

I know how to do this! it's just been a while, it's a review :/

Answer 4

I still need help!

Answer 5

You can use the quadratic formula for this..
x = -b +- √b^2 - 4ac / 2a

Answer 6

I think doing that would be easier than factoring it out, since the solutions are complex.

Answer 7

That formula doesn't look very familiar. Would I just plug in the numbers?

Answer 8

ax^2 + bx + c = 0
those are the values of a b and c.
have you taken algebra 1? you learn the quadratic formula then..

Answer 9

I'm in Algebra 2! I'm really struggling though, and i have a hard time remembering formulas and steps. I'm completely familiar with ax^2 + bx + c = 0. And a few weeks ago we were learning about quadratic formulas, but what you showed me doesn't look familiar. Maybe we learned it a different way.

Answer 10

\(\large ax^2+bx+c=0\)


quadratic formula to find solutions :-

\(\large x~=~\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

Answer 11

3(4x^2 + 4x + 5) = 0

\(\large 4x^2 + 4x + 5 = 0\)

\(a = 4\)
\(b = 4\)
\(c = 5\)

plug them above

Answer 12

okay!! I think I understand that. But where does the distributive 3 go?

Answer 13

that goes away

Answer 14

\(3(4x^2 + 4x + 5) = 0\)

\(3 \ne 0\),

so,

\(4x^2 + 4x + 5 = 0\)

Answer 15

look at below example :-

\(100x = 0\)

\(100 \ne 0\)

so,

\(x = 0\)

where did the 100 go ?

Answer 16

because 100 does not equal 0, it just simply disappears?

Answer 17

since 100 does not equal 0, the other guy must equal 0

Answer 18

would the square root of -64 be 8i?

Answer 19

yup

Answer 20

and -4 x 8i would be.. -32i?

Answer 21

careful, its a + or - sign in between
not multiplication

Answer 22

Oh! it would be addition, right?

Answer 23

-4 \(\pm\) 8i

Answer 24

so -4 + 8i = 4i?

Answer 25

and then 4i divided by 8?

Answer 26

noo

Answer 27

:(

Answer 28

-4 + 8i
-4 - 8i
those are your two answers ._.

Answer 29

i'm hopeless.

Answer 30

divide by 2a

Answer 31

i think i'm over thinking it.

Answer 32

Well a is 4, right? and 2 times 4 is 8!

Answer 33

-4 +- 8i / 2(4)
=
-4 +- 8i / 8

Answer 34

\(\large ax^2+bx+c=0\)

quadratic formula to find solutions :-

\(\large x~=~\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\large 4x^2 + 4x + 5\)

\(\large x~=~\frac{-4\pm\sqrt{4^2-4 \times 4 \times 5}}{2 \times 4}\)

\(\large x~=~\frac{-4\pm\sqrt{16-80}}{8}\)

\(\large x~=~\frac{-4\pm\sqrt{-64}}{8}\)

\(\large x~=~\frac{-4\pm 8i}{8}\)

\(\large x~=~\frac{-1\pm 2i}{2}\)

Answer 35

Ohhh, okay! So we're simplifying for the most part. Not solving.

Answer 36

dint get u

Answer 37

thanks, ganshie, its too early in the morning for me to do quadratics and what not xD

Answer 38

np :)

Answer 39

I'm afraid to guess what the solutions would be.. it would be -1 +2i and -1 - 2i, right?

Answer 40

who ate the 2 in bottom ?

Answer 41

\(\large ax^2+bx+c=0\)

quadratic formula to find solutions :-

\(\large x~=~\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\large 4x^2 + 4x + 5\)

\(\large x~=~\frac{-4\pm\sqrt{4^2-4 \times 4 \times 5}}{2 \times 4}\)

\(\large x~=~\frac{-4\pm\sqrt{16-80}}{8}\)

\(\large x~=~\frac{-4\pm\sqrt{-64}}{8}\)

\(\large x~=~\frac{-4\pm 8i}{8}\)

\(\large x~=~\frac{-1\pm 2i}{2}\)

\(\large x~=~\frac{-1 + 2i}{2} ~~or ~~ x~=~\frac{-1 - 2i}{2} \)

Answer 42

I've never seen it like that before. I would keep the denominator?

Answer 43

yup :)

Answer 44

okay wow! that's what's completely throwing me off!

Answer 45

\(\large ax^2+bx+c=0\)

quadratic formula to find solutions :-

\(\large x~=~\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

\(\large 4x^2 + 4x + 5\)

\(\large x~=~\frac{-4\pm\sqrt{4^2-4 \times 4 \times 5}}{2 \times 4}\)

\(\large x~=~\frac{-4\pm\sqrt{16-80}}{8}\)

\(\large x~=~\frac{-4\pm\sqrt{-64}}{8}\)

\(\large x~=~\frac{-4\pm 8i}{8}\)

\(\large x~=~\frac{-1\pm 2i}{2}\)

\(\large x~=~\frac{-1 + 2i}{2}~~ or~~ x~=~\frac{-1 - 2i}{2} \)

\(\large x~=~\frac{-1}{2} + i ~~ or~~ x~=~\frac{-1}{2} - i \)

Answer 46

you may leave it like this also

Answer 47

I like the other one better. :) but thank you so much!!! i can't believe you actually put up with me! haha. I really appreciate it.

Answer 48

lol you're good its fine :)