SOLUTION: Good afternoon. I need some help with this problem. This is under the title of solving polynomial equations. Solve (x+3)*(x-3)= 40. Identify any double roots. I've tried d

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Good afternoon. I need some help with this problem. This is under the title of solving polynomial equations. Solve (x+3)*(x-3)= 40. Identify any double roots. I've tried d      Log On


   

Question 256939: Good afternoon. I need some help with this problem.
This is under the title of solving polynomial equations.
Solve (x+3)*(x-3)= 40. Identify any double roots.
I've tried different methods
(x+3)*(x-3)=40 = x^2-9=40 = Answer {7,-7}

(x+3)*(x+3)+40=0
40(x+3)*(x-3)=0
40(x^2-9)=0
5(x-3)*8(x+3)=0
Answer: {3, -3}
or
(x+3)*(x-3)=40
(7+3)*(-7-3)=40
-70=40
Answer {0,3,-3}
I'm really confused so any help is welcome thank you.
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the original equation
%28x%2B3%29%2A%28x-3%29=+40
step 1 - foil the left side to get
x%5E2+-+9+=+40
step 2 - add 9 to get
x%5E2+=+49
step 3 - take a square root to get
x=7, x=-7
In this case there are no double roots.
--
to check let x = 7 and we get
(7+3)(7-3) = 40 - - - > 10*4 = 40 = 40 [true]
Now let x = -7
(-7+3)(-7-3) = 40 - - > (-4)(-10) = 40 = 40 [true]