Question 916822: I really need help understanding this ;/
What are the zero(s) of the function f(x) = the quantity of 5 x squared minus 25 x, all over x?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! equation is y = (5x^2 + x)/x
set y = 0 to get:
(5x^2 + x)/x = 0
multiply both sides of the equation by x to get:
5x^2 + x = 0
this is a quadratic equation that you can factor using any of the methods that work.
i'll use the quadratic formula.
since the eqaution is in standard form of ax^2 + bx + c = 0, you can get the values of a, b, and c as follows:
a = 5
b = 1
c = 0
the quadratic formula is x = (-b +/- sqrt(b^2-4ac))/(2a)
replace a with 5 and b with 1 and c with 0 and you will get:
x = 0 or x = -1/5
those are your possible solutions.
you have to confirm that they are good by replacing x with them in the original equation to see if the equation is true.
always confirm using the original equation or expression.
the original equation is (5x^2 + x)/x = 0
when x = 0, the original equation of (5x^2 + x)/x = 0 becomes (0/0) = 0 which becomes undefined = 0 which is not true.
the expression 0/0 is undefined because the denominator is equal to 0.
x = 0 is therefore not a solution to this problem.
when x = -1/5, the original equation of (5x^2 + x)/x = 0 becomes (5*(1/25 - 1/5) / (1/5) = 0 which becomes (1/5 - 1/5) / (1/5) = 0 which becomes 0 / (1/5) = 0 which becomes 0 = 0 which is true.
x = -1/5 is therefore the only solution to this problem.
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