Question

Find all real zeros of the equation y = x + 1/(x+2) + 4

Answer 1

\[y=x+\frac{1}{x+2}+4\] correct?
We can make a common denominator (x+2). We multiply each term by (x+2)/(x+2):
\[y=\frac{x(x+2)}{x+2}+\frac{1}{x+2}+\frac{4(x+2)}{x+2}\]
\[y=\frac{x^2+2x+1+4x+8}{(x+2)^3}=\frac{x^2+6x+9}{(x+2)^3}\]
Set the denominator equal to 0 to find the excluded values (you can't divide by 0).
(x+2)^3=0 When x=-2, it becomes (-2+2)^3=(0)^3=0. x=/=-2.

Now, set the numerator equal to zero to find the solutions.
x^2+6x+9=0 We can factor this. x^2+3x+3x+9=0 | x(x+3) +3(x+3)=0 | (x+3)(x+3)=0 | (x+3)^2=0
We can solve this easily now. x=-3, because (-3+3)^2=(0)^2=0.

Now we check this solution. Is it one of our excluded values?
x = -3
x =/= -2
No, so the solution is -3.