Question

if 1byx - 1byx-2 = 3, xnotequal 0, 2 find value x ​

Answer 1

Step-by-step explanation:

We have

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

1 1

--- - ------ = 3

x x-2

The common denominator is x(x-2)

1 x-2 1 x

---• ----- - ----- • --- = 3

x x-2 x-2 x

x-2 x

------- - ------- = 3

x(x-2) x(x-2)

x-2 - x

-------- = 3

x(x-2)

-2 = 3x(x-2)

-2 = 3x2-6x

3x2-6x+2 = 0

Using the quadratic formula we can solve for x

-b±√(b2-4ac)

--------------

2a

6±√[62-4(3)(2)]

-------------------

2(3)

6±√(36-24)

-------------

6

6±√12

-------

6

6±2√3

-------

6

3±√3

------

3

Answer 2

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  $x =\frac{3+1\sqrt{3}}{3}$,$x =\frac{3-1\sqrt{3}}{3}$ .

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