Question

((x^2)+4x/x-5)>0

What is the set of solutions to this problem?

Answer 1

is that x-5 the denominator like this?
\[\large x^2+\frac{4x}{x-5} \]>0

Answer 2

no, (x^2)+4x is the numerator. and then x-5 is the denominator.

Answer 3

so it's like this...
\[\large \frac{x^2+4x}{x-5}>0 \]

Answer 4

yup ^^

Answer 5

first we need to solve this...
\[\large \frac{x^2+4x}{x-5}=0 \]

Answer 6

so we need to factor the top...
\[\large \frac{x(x+4)}{x-5}=0 \]
this implies x(x-4)=0 so x= 0, x=-4
correct?

Answer 7

sorry i mean x(x+4)=0 so x=0, x=-4

Answer 8

these are the zeros (where the graph crosses the x axis) of the expression on the left hand side.

Answer 9

yeah, i got that too. but then the question asks us to state the solution using interval notation.
so will it be like:
(-infinity,-4)U(0,infinity)?
cuz i tried that.. and it was wrong ><

Answer 10

yes... but we need to consider 1 more thing... that is the vertical asymptote... to get the vertical asymptote, you need to set the denominator equal to zero... then solve for x. what x value will make the denominator equal to zero?

Answer 11

5

Answer 12

ok good... now we must check intervals...