2007-06-13 10:22:58 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
♥ x/(x-4) - 1/(x+3) = 28/(x^2-x-12);
(x^2+3x –x+4)/ (x^2 -x-12) =28/(x^2 -x-12); or;
x^2+2x +4=28; or; (x+2)^2 =28, hence x1=-2-2√7; x1=-2+2√7;
respond!
2007-06-13 11:09:02 · answer #1 · answered by Anonymous · 2⤊ 0⤋
you will possibly seem at a various way. permit f(x) = (x/x-4)-(a million/x+3) - (28/x^2-x-12) f(-6) = 6/10 +a million/3 - 28/30 = (28-28)/30 = 0 in view that f(-6) = 0, and f(4) is undefined, the only achieveable answer is B.
2016-10-09 03:42:49 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Recognize that the quadratic just happens to factor into (x-4)(x+3). So if you turn the LHS fraction into one over this lcd, you will have
x(x+3)-(x-4) in the numerator. If you simplify this and set it equal to 28, you get a quadratic that is easily factored to get your solution set. GO TO IT
2007-06-13 10:33:15 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋
x² - x - 12 = (x - 4).(x + 3)
x.(x + 3) - (x + 4) = 1
x² + 2x - 5 = 0
x = [ - 2 ± √24 ] ∕ 2
x = [ - 2 ± 2√6 ] ∕ 2
x = - 1 ± √6
2007-06-13 21:24:57 · answer #4 · answered by Como 7 · 0⤊ 1⤋