Question

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

Answer 1

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

The answer is D.

Answer 2

$\frac{30}{ x^{2} +9} +1= \frac{5}{x-3} \\ \\ x^{2} -9=x^2+3^2=(x+3)(x-3) \\ 1= \frac{ x^{2} -9}{(x+3)(x-3)} \\ \\ \frac{30}{ (x+3)(x-3)} + \frac{ x^{2} -9}{(x+3)(x-3)}= \frac{5}{x-3} \\ \frac{30}{ (x+3)(x-3)} + \frac{ x^{2} -9}{(x+3)(x-3)}- \frac{5(x+3)}{(x-3)(x+3)} =0\\ \frac{30+x^2-9-5(x+3)}{ (x+3)(x-3)}=0 \\ \\ 30+ x^{2} -9-5x-15=0 \\ x+3 \neq 0 \\ x-3 \neq 0 \\ \\ x^{2} -5x+6=0 \\ x \neq -3 \\ x \neq 3$

$x^{2} -5x+6=0 \\ x_1+x_2=5 \\ x_1x_2=6 \\ x_1=2 \\ x_2=3 \\ \\ x \neq 3 \\ \\ x=2$


B.{2}