Question

Use complex conjugates to factor the expressions.

3x^2 + 12

Answer 1

Need help with this! Will medal :)

Answer 2

Factor 3 out of 3x^(2) + 12.
3(x^(2)+4)

Answer 3

Given: 3x^2+12
1) Factor the '3' out of each term.
Hanna has beaten me, so I'll put you into her good hands. ;)

Answer 4

Hanna, would you like to finish helping with this problem? If so, please proceed.

Answer 5

oh no im sorry you can do it you would have the right answer and i would not want to give him the wrong one

Answer 6

@mathmale

Answer 7

Hanna is correct here: " 3(x^(2)+4) "

For now, ignore the factor 3.

That leaves us with x^2+4.

We could cheat (just a tiny, little bit) by applying the quadratic formula here. You remember that, don't you? x^2+4 is the same as 1x^2 + 0x + 4.

What are "a," "b" and "c" here?

Answer 8

Noodles: Offline already? Please, would you please not post and disappear.

Answer 9

Sorry I'm in class, I have my midterm coming up so I had to hurry and learn this before! Appreciate the help :)

Answer 10

And actually Quadratic formula is one of my weak points, that's probably why I don't understand this haha

Answer 11

Want help or not?

Answer 12

Yes I need it!

Answer 13

compare your " x^2+4 " (which is the same as 1x^2 + 0x + 4) to
ax^2 +bx + c.

What are a, b and c?

Glad you've studied the q. formula before. Note that review could be as simiple as Googling "quadratic formula."

Answer 14

a=?
b=?
c=?

Answer 15

A=1
B=0
C=4

Is it necessary to put a 1 in front of x?

Answer 16

I wanted you to understand where the value of 'a' can be found. x^2 and 1x^2 are equivalent, but if you want the coeff. 'a' then please write the quadratic as ax^2+bx+c.

Answer 17

Here's the quadratic formula. Hope you'll write it down now if you haven't already\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 18

Oh wow mine looked a bit different... lol

Answer 19

This formula will give you the exact roots of the given quadratic equation.

This info is important, because it will enable you to write the factors of the given equation quickly.

Answer 20

Do I plug in the values for a b and c?

Answer 21

Yes.

Answer 22

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

Answer 23

Here, a=1, b=0 and c=4

Answer 24

so x = ?

Answer 25

Note: this is a quadratic (2nd degree polynomial), so you must obtain 2 roots.

Answer 26

I'm gonna try to work it real quick, I'll let you know if I have any problems

Answer 27

How's your solution coming along?

Answer 28

I multiplied by 2 to get rid of the fraction, Im lost after that

Answer 29

\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
with b=0, a=1 and c=0

becomes \[x=\frac{ \pm \sqrt{-4(1)(4)} }{ 2 }\]

Answer 30

Do you agree or disagree with this?

No need to multiply by 2 (or to eliminate "2" from the denom.

Answer 31

You do end up multiplying 4 by a and c to get 16 yes?

Answer 32

Yes, but you'll have a neg. number under the radical sign.

Write out your equation for x, please.

Answer 33

\[x=\frac{ \pm \sqrt{-4(1)(4)} }{ 2 }\]

Answer 34

Simplify this.

Answer 35

I can't write it on my phone, but all I did was multiply for 16, I'm not sure what to do after that.

Answer 36

\[x=\frac{ \pm \sqrt{-4(1)(4)} }{ 2 }=\frac{ \pm \sqrt{-4} }{ 2 }\]

Answer 37

What do you know about the square root of a negative number?

Tell me your result(s) in words, please.

Answer 38

It is in terms of i yes? For the square root of a negative?

Answer 39

The "discriminant" is " b^2 - 4 a c "

Here b=0, so we have - 4 a c

a is 1 and c is 4. So, what is - 4 a c ?

Answer 40

generic: are we still working on this problem?

If a=1 and c=4, what is - 4 a c ?

Answer 41

This is the 2nd time you have disappeared without notice. Please let your helper(s) know when you must log off. Thank you.

Answer 42

Lol you could stop having an attitude and just put what I need to know here for when I come back, I'll see it anyway. I'm not asking for answers, just the steps to get it.