Algebra Help?

Question

how do you solve the problem

x squared +6x+9=0

please show step by step.

thank you very much!

Answer 1

factoring gives you:
(x + 3)^2 = 0

x = -3

Answer 2

x^2 + 6x + 9 = 0
(x + 3)(x + 3) = 0
x + 3 = 0, x = - 3

Answer: x = - 3

Proof:
- 3^2 + 6(- 3) + 9 = 0
9 - 18 + 9 = 0
0 = 0

Answer 3

x^2+6x+9=0
Factor the equation into
(x+3)^2=0 or (x+3)(x+3)=9
Set it equal to zero
x+3=0 They are the same so only need to do once
Solve
x=-3

p.s. to show squared, cubed, etc just use ^ and whatever number the exponent is.

Answer 4

x^2+6x+9=0
You have to factor this in order to find the zeros. so you get
(x+3)(x+3)=0
You have to find both numbers that add to get 6 but multiply to get 9. That number is 3.
So you solve for zero with each of the factors.
(x+3)=0
x=-3
Because both factors are the same, you do not need to do it twice.
Hope this helps.
~Scott~

Answer 5

you have to factor it...find 2 numbers that added give you 6 and multiplied give you 9...that is how factoring works...i'll show you...

x^2+6x+9=0

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

because 3+3=6
and (+3)x(+3)=9

after this step,you have to find the values that make x=0...and they are -3 and -3...

so the answer is

x=-3

clear now?

Answer 6

x^2+6x+9=0
-9 -9
x^2+6x= -9
-6 -6
x^2+x= -3
-x -x
x=-3

Answer 7

x^2+6x+9=0
can be solved 2 ways
Way#1
This is a perfect square(I know this from experience.You will too, in the near future)
(x+3)(x+3)=0, x=-3

Way#2
Quadratic Formula, x={-b+/-rt(b^2-4ac)} / 2a
a= coefficient of x^2 term, here+1
b= coefficient of x term, here +6
c= constant term, here +9
x={ -6+/- rt(36-4(1)(9))}/2
x={-6+rt(36-36)}/2
x=-6/2,
x=-3

Answer 8

x squared +6x+9=0

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

x+3=0

x=-3

There two same roots.