Question

Solve: 1/x - 1/x-3=3, where x is not equal to zero​

Answer 1

Answer:

$\frac{1}{x} - \frac{1}{x - 3} = 3 \\ \therefore \: \frac{x - 3 - x}{x(x - 3)} = 3 \\ \therefore \: \frac{ - 3}{x(x - 3)} = 3 \\ \therefore \frac{ - 1}{x(x - 3)} = 1 \\ \therefore \: - 1 = x(x - 3) \\ \therefore \: 0 = {x}^{2} - 3x + 1 \\ \therefore \: {x}^{2} - 3x + 1 = 0 \\ comparing \: it \: with \: a {x}^{2} + bx + c = 0 \\ we \: find : \\ a = 1 \\ b = - 3 \\ c = 1 \\now \\ {b}^{2} - 4ac \\ = {( - 3)}^{2} - 4 \times 1 \times 1 \\ = 9 - 4 \\ = 5 \\ by \: formula \\ x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a} \\ = \frac{ - ( - 3)\pm \sqrt{ 5 }}{2 \times 1} \\ \therefore \: x \: = \frac{ 3\pm \sqrt{ 5 }}{2 } \\ so \\ x =\frac{ 3 + \sqrt{ 5 }}{2 } \: \: or \: \: x =\frac{ 3 - \sqrt{ 5 }}{2 }$