Solving (f+g)(x)=0 using f(x) and g(x)

[Howdoyoumath]

I have another question that I don't understand.
f(x)=x^2−6x and g(x)=x+6, find
(f+g)(x)=0
(f-g)(x)=12
(fg)(x)=-36
(f/g)(x)=-1
It says my answers are incorrect and i plugged in f(6)=(6)^2-6(6) and for g(-6)=(-6)+6
f(6)=0 g(-6)=0
Where did I go wrong? Also should they always =0? Also, should f and g have the same f(6) or g(6) value?

 

[Steven G]

Let's see where you went wrong.
(f+g)(6) = f(6) + g(6) = 0 + 12 = 12, NOT 0
(f+g)(-6) = f(-6) + g(-6) = 72 + 0 = 72, NOT 0
Just because f(6) = 0 does NOT imply that (f+g)(6) = 0
You first have to find (f+g)(x) and figure out when that equals 0

 

[Otis]

… i plugged in f(6) … and for g(-6) …

Howdy, Howdo… You solved f(x)=0 and g(x)=0 thinking that (f+g)(x) would then be 0+0. It doesn't work that way; for one reason, the variable x cannot represent two different values (6 and -6) at the same time. Here's how to proceed:

(f+g)(x) = 0

We know the left-hand side is f(x)+g(x), so we add the given expressions for f(x) and g(x) together. That will give us the expression for (f+g)(x).

x^2 − 6x + x + 6 = 0

That is the equation you need to solve, to find all values of x for which the equation (f+g)(x)=0 is true.

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