How to do this math equation?

Question

4x^2+12x+9=0

Solve for x...

Include how you did it please.

I think it has something to do with factoring...

Answer 1

Factorize;

You know the two values for x multiply to equal 4, (4x^2), so it's either 2, 2 or 4, 1.

You also know that the numbers being added or taken from the value for x multiply to make positive 9. This means that it could be 9 and 1, or 3 and 3, either both positive or both negative.

You also know that if you multiply the number being added to the value of x in the opposite bracket with both numbers, you get 12x.

Because of this, you can tell that only the below would work.

(2x + 3)(2x + 3) = 0

So one, or both of the brackets has to equal 0 in order for it to multiply to  equal 0;

if 2x + 3 = 0

2x = -3

x = -1.5

Answer 2

4x^2 + 6x + 6x + 12 = 0

2x( 2x + 3) + 3(2x + 3) = 0

(2x + 3)(2x + 3) = 0

(2x + 3)^2 = 0

x =  -3/2

Answer 3

1) Use the box method on the left side of the equation (factor it out), giving you: (2x+3)(2x+3)=0

2) For that equation to be true, one of those sets of parenthesis has to be = to 0, since they're the same, you just have to prove one.

3) 2x+3=0, subtract 3

4) 2x = -3

5) Divide by 2

6) X = -3/2

Answer 4

-3.11

repeating

Answer 5

(2x+3)^2 = 0

x=-1.5

Answer 6

You can factor if you want.

But the sure fire way is to use the quadratic equation.  That will get you the answers anytime.  Just look it up if you don't know it.

If you want to factor it, then it's just a bunch of trial and error until you get it right.

First start with the two prenthases (x+ #)(x+ #).

Then you have to decided whether to do 4x and x or 2x and 2x (to multiply to give you 4x^2.

After that you have to play with the two number so that when you multiply it out it gives you 9 (which pretty much leaves you with +/- 1 and 9, or +/- 3).

You just have to work it out so that it adds up right.  Keep playing with it until you get it right.

Answer 7

You have to factor this out.

multiples of 9 can only be 9,1 or 3,3. So since you also have a 4 to factor, you will add up to 12.

Since all symbols are positive, you know both factored equations will be "+"

(2x+3)(2x+3)=0

so (2x+3)^2=0

so 2x+3=0

so 2x=-3

x=-1.5

Answer 8

factor using decomposition.

multiply the first constant and the last constant to obtain 36.

figure out numbers that add to 12, and multiply to 36. so +6 and +6

new equation...

4x^2 + 6x + 6x + 9 = 0 simplifying,

2x(x + 3) + 3(x + 3) = 0

(2x + 3) , (x + 3) = 0

thus x = -3, -3/2

Answer 9

Yes, it does.

First, use the quadratic formula to see if you can get any integers [[don't want any imaginary numbers]]

but it's very simple so this is what I got:

(2x+3)(2x+3)=0

a.k.a (2x+3)^2 because it's a perfect!

so then the anser would be -3/2

Answer 10

when did they start putting letters in there?

Answer 11

factor using decomposition.

multiply the first constant and the last constant to obtain 36.

figure out numbers that add to 12, and multiply to 36. so +6 and +6

new equation

4x^2 + 6x + 6x + 9 = 0 simplifying,

2x(x + 3) + 3(x + 3) = 0

(2x + 3) , (x + 3) = 0

thus x = -3, -3/2

Answer 12

yes you factor. but if its unfactorable then you use the quadratic equation to solve for the values of x.

like this just plug in your numbers..

     ax^2+bx=c then plug it into...

       {-b+(b^2-4ac)^1/2}/2a and {-b-(b^2-4ac)^1/2}/2a to get the other answer beacuse there should be 2 solutions