Question 1171029: Just simply rewrite the given equation to its standard form of quadratic equation then find the value of a, b and c
x/(x-5) - 3/(x+1) = 30/(x2-4x-5)
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! standard form of the equation is ax^2 + bx + c = 0
a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.
to get the equation into this form, subtract 30/(x^2-4x-5) from both sides to get:
x/(x-5)-3/(x+1)-30/(x^2-4x-5) = 0
multiply both sides of the equaton by x^2-4x-5 to get:
x*(x^2-4x-5)/(x-5) - 3*(x^2-4x-5)/(x+1) - 30 = 0
if you factor x^2-4x-5, you get (x-5)*(x+1).
replace x^2-4x-5 with (x-5)*(x+1) to get:
x*(x-5)*(x+1)/(x-5) - 3*(x-5)*(x+1)/(x+1) - 30 = 0
simplify to get:
x*(x+1) - 3*(x-5) - 30 = 0
simplify further to get:
x^2 + x - 3x + 15 - 30 = 0
combine like terms to get:
x^2 - 2x - 15 = 0.
the equation is now in standard form, where:
a = 1
b = -2
c = -15
what i found when i solved for x in both equaion was:
for x^2 - 2x - 15, y = 0 when x = 5 or x = -3
for x/(x-5)-3/(x+1)-30/(x^2-4x-5), y = 0 when x = -3.
when x = 5, the denominator in the equation became 0, making x = 5 not a solution to this equation.
since you have to go back to the original equation to test out your solution, it appears that the only solution for the original equation is x = -3, even though x = 5 was a solution for the modified equation.
you were not, however, asked to find a solution.
you were only asked to rewrite the equation to make it in the standard form of a quadratic equation.
i believe i did that correctly, getting you:
x^2 - 2x - 15 = 0, where:
a = 1
b = -2
c = -15
i believe that's your solution.
i couldn't see any other way to get it into standard form.
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Answer by MathTherapy(10552)  (Show Source):
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