Question

Solve the given quadratic equation by factorization :

$\frac{1}{x} - \frac{1}{x - 2} = 3$
, x ≠ 0 , 2​

Answer 1

Step-by-step explanation:

1/x - 1/x-2 = 3

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

x-2-x/x²-2x=3

-2=3x²-6x

3x²-6x+2 =0

Using the quadratic formula we can solve for x

-b±√(b2-4ac)

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2a

6±√[62-4(3)(2)]

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2(3)

6±√(36-24)

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6

6±√12

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6

3±√3

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3

Answer 2

Answer:

Required value of x is either ( 3 + √3 ) / 3 or ( 3 - √3 ) / 3 .

Step-by-step explanation:

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

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

= > [ x - 2 - x ] / ( x^2 - 2x ) = 3

= > - 2 = 3( x^2 - 2x )

= > 3x^2 - 6x + 2 = 0

= > 3x^2 - 6x - √3 + √3 + 2 = 0 { Add and subtract √3 }

= > 3x^2 + ( - 3 + √3 - 3 - √3 )x - √3 + √3 + ( 3 - 1 ) = 0 { - 6 = - 3 - 3 or - 3 + √3 - 3 - √3 and 2 = 3 - 1 }

= > 3x^2 - 3x + √3 x - 3x - √3 x - √3 + √3 + 3 - 1 = 0

= > 3x^2 - 3x + √3 x - 3x + 3 - √3 - √3 x + √3 - 1 = 0

= > √3 x( √3 x - √3 + 1 ) - √3 ( √3 x - √3 + 1 ) - ( √ 3 x - √3 + 1 ) = 0

= > ( √3 x - √3 x + 1 )( √3 x - √3 - 1 ) = 0

= > [ √3( √3 x - √3 - 1 ) ][ √3( √3 x - √3 + 1 ) ] = 0 { multiplied by 3 or √3 x √3 }

= > ( 3x - 3 - √3 )( 3x - 3 + √3 ) = 0

= > 3x = 3 + √3 Or 3x = 3 - √3

= > x = ( 3 + √3 ) / 3 Or x = ( 3 - √3 ) / 3

Thus,

Required value of x is either ( 3 + √3 ) / 3 or ( 3 - √3 ) / 3 .