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Roots are x=-5, x=3, the x=3 is an extraneous root.
ok thanks. could you show me how you found those answers?
Yes be glad to. First I did the addition. The common denominator is (x+2)(x-3)
The numerators are then x(x-3) +5(x+2) over the common denominator of (x+2)(x-3) kind of like this: x(x-3) +5(x+2) -------------- =25/(x^2-x-6) (x+2)(x-3) You follow so far?
yes i do.
Now lets work on the denominator of the fraction on the right side of the equal sign. the denominator is x^2-x-6 This can be factored to look like (x-3)(x+2) Now we have on the right of the equal sign: 25 ------------------ (x-3)(x+2) You still following?
Now look at both fractions, notice the denominator on the left is also equal to the denominator on the right of our two fractions that are equal. Do you see that?
yes i do
well since they are equal, then for the two fractions to be equal, the numerators must be equal,so we now can write: x(x-3) + 5(x+2) = 25 Do you follow that kind of reasoning?
ok i see
can you take it from there are do you want me walk you thru the solution.?
no i can take it from here. thank you very much!
O.K and the root equal 3 is extraneous as it results in a division by zero and as you know that is a no-no
ok, so the final equation before you get (x+5)(x-3) is x^2+2x-15 correct?