Question

Solve the equation.

Answer 1

\(x^2+12x+36=25\)

Answer 2

try subtracting 25 on both sides first
so you have something in the form ax^2+bx+c=0
and you could factor ax^2+bx+c or use the quadratic formula to give you x for when ax^2+bx+c=0

Answer 3

okay :) \(x^2+12x+11\)?

Answer 4

And x^2+12x+11 is actually factorable if you want to factor it

Answer 5

ummm (x+1)(x+11)? i don't know

Answer 6

I think that's it

Answer 7

\[x^2+12x+36=25 \\ x^2+12x+36-25=23-25 \\ x^2+12x+11=0 \\ (x+1)(x+11)=0\]

Answer 8

okay

Answer 9

yes because 11(1)=11 and 11+1=12

Answer 10

and that second line I hip you notice 23 is suppose to be 25 :p

Answer 11

hope* lol

Answer 12

:) yes I did :) what do I do next?

Answer 13

\[x^2+12x+36=25 \\ x^2+12x+36-25=25-25 \\ x^2+12x+11=0 \\ (x+1)(x+11)=0\]

Answer 14

Now since you have a*b=0 then either a=0 or b=0 or both =0

Answer 15

huh?

Answer 16

how do I know that?

Answer 17

so you solve both x+1=0 and solve also x+11=0

\[x^2+12x+36=25 \\ x^2+12x+36-25=25-25 \\ x^2+12x+11=0 \\ (x+1)(x+11)=0 \\x+1=0 \text{ or } x+11=0\]

Answer 18

if a*b=0 then a=0 or b=0 or both =0
because 5*0=0 and 0*2=0 and 0*0=0

Answer 19

at least one of those factors have to be 0

Answer 20

oh yeah......:)

Answer 21

so what's your two solutions?

Answer 22

x=-1 x=-11???

Answer 23

i have no idea actually

Answer 24

yep x+1=0 when x=-1
and x+11=0 when x=-11

Answer 25

okay

Answer 26

You would state the solution as x=-1 or x=-11

Answer 27

or \[x \in \left\{ -1,-11 \right\}\]

Answer 28

okay....what do I do with that?

Answer 29

Nothing. It is just stating the solution set

Answer 30

It is the set of solutions

Answer 31

okay thanks! :)