Question

find roots of quadratic equation 1/x - 1/x-2 = 3

Answer 1

Answer:

The roots of the quadratic equation $\frac{1}{x}-\frac{1}{x-2} = 3$ are  $x =\frac{3+1\sqrt{3}}{3}$,$x =\frac{3-1\sqrt{3}}{3}$ .

Step-by-step explanation:

As given the equation in the form

$\frac{1}{x}-\frac{1}{x-2} = 3$

Simplify the above equation

(x-2)-x = 3x × (x-2)

x-2 - x = 3x² - 6x

3x² - 6x + 2 = 0          

As the equation is written in the form ax² + bx + c = 0

$x =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

a = 3 , b = -6 , c = 2

Put all the values in the above equation

$x =\frac{-(-6)\pm\sqrt{(-6)^{2}-4\times 3\times 2}}{2\times 3}$

$x =\frac{6\pm\sqrt{36-24}}{6}$

$x =\frac{6\pm\sqrt{12}}{6}$

$x =\frac{6\pm2\sqrt{3}}{6}$

$x =\frac{3\pm1\sqrt{3}}{3}$

Thus

$x =\frac{3+1\sqrt{3}}{3}$

$x =\frac{3-1\sqrt{3}}{3}$

Therefore the roots of the quadratic equation $\frac{1}{x}-\frac{1}{x-2} = 3$ are  $x =\frac{3+1\sqrt{3}}{3}$,$x =\frac{3-1\sqrt{3}}{3}$ .

Answer 2

Answer:

x =3+1√3/3

x=3-1√3/3

Step-by-step explanation: