Question

Find the roots of 1/x - 1/x-2 =3

Answer 1

Answer:

Step-by-step explanation:

1/x - 1/(x-2) = 3

taking LCM as (x)(x-2)

we get

(x-2- x) / (x(x-2)) = 3

-2 = 3x^2 -6x

taking all terms to one side we get

3x^2 - 6x + 2 = 0

Now, we can solve this quadratic equation using the formula

(-b (+-) $\sqrt{b^{2} -4ac }$)/2a

6 (+-) 2$\sqrt{3}$/6 = 1(+-) 1/$\sqrt{3}$

therefore your roots are

1 + 1/$\sqrt{3}$ and 1 - 1/$\sqrt{3}$

CHEERS MATE

PLEASE MARK BRAINLIEST

Answer 2

$\frac{1}{x} - \frac{1}{x - 2} = 3 \\ \frac{x - 2 - x}{x(x - 2)} = 3 \\ \frac{ - 2}{ {x}^{2} - 2x } = 3 \\ 3 {x}^{2} - 6x = - 2 \\ 3 {x}^{2} - 6x + 2 = 0$

Comparing the given polynomial with ax²+bx+c = 0, we get

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

We know

D= b²-4ac

= (-6)²-4(3)(2)

= 36-24

= 12

Here, D> 0

Therefore, the given quadratic polynomial has two distinct real roots.

Using quadratic formula

x= (-b±√D)/2a

x= (6±√12)/2×3

x= (6±2√3)/6

x= (3±√3)/3

x = (3+√3)/3, (3-√3)/3