Question

What is the solution set of the equation 30/x^2-9 +1=5/x-3

A. {2,3}
B.{2}
C.{3}
D.{ }

Answer 1

`(30)/( x^(2) +9) +1= (5)/(x-3) \\ \\ x^(2) -9=x^2+3^2=(x+3)(x-3) \\ 1= ( x^(2) -9)/((x+3)(x-3)) \\ \\ (30)/( (x+3)(x-3)) + ( x^(2) -9)/((x+3)(x-3))= (5)/(x-3) \\ (30)/( (x+3)(x-3)) + ( x^(2) -9)/((x+3)(x-3))- (5(x+3))/((x-3)(x+3)) =0\\ (30+x^2-9-5(x+3))/( (x+3)(x-3))=0 \\ \\ 30+ x^(2) -9-5x-15=0 \\ x+3 \\eq 0 \\ x-3 \\eq 0 \\ \\ x^(2) -5x+6=0 \\ x \\eq -3 \\ x \\eq 3 \\`


`x^(2) -5x+6=0 \\ x_1+x_2=5 \\ x_1x_2=6 \\ x_1=2 \\ x_2=3 \\ \\ x \\eq 3 \\ \\ x=2`


B.{2}

Answer 2

`(30)/(x^(2) - 9 + 1) = (5)/(x - 3) \\(30)/(x^(2) - 8) = (5)/(x - 3) \\5(x^(2) - 8) = 30(x - 3) \\5(x^(2)) - 5(8) = 30(x) - 30(3) \\5x^(2) - 40 = 30x - 90 \\5x^(2) + 50 = 30x \\5x^(2) - 30x + 50 = 0 \\5(x^(2)) - 5(6x) + 5(10) = 0 \\5(x^(2) - 6x + 10) = 0 \\(5(x^(2) - 6x + 10))/(5) = (0)/(5) \\x^(2) - 6x + 10 = 0 \\x = \frac{-(-6) \± \sqrt{(-6)^(2) - 4(1)(10)}}{2(1)} \\x = (6 \± √(36 - 40))/(2) \\x = (6 \± √(-4))/(2) \\x = (6 \± 2i)/(2) \\x = 3 \± i \\x = 3 + i\ or\ x = 3 - i`

The answer is D.