Question

Solve the following quadratic equations by factorization: $\frac{1}{(x-1)(x-2)} +\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}= \frac{1}{6}$

Answer 1

it is the answer...plz mark me as brainliest...

Answer 2

Solution :

i ) 1/[(x-1)(x-2)]

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

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

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

Similarly ,

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

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

iii ) 1/[(x-3)(x-4)]

= 1/(x-4) - 1/(x-3) ----( 3 )

Now ,

( 1 ) + ( 2 ) + ( 3 ) = 1/6

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

+ 1/(x-4) - 1/(x-3) = 1/6

After cancellation , we get

=> -1/(x-1) + 1/(x-4) = 1/6

=> [ -(x-4) + (x-1) ]/[ (x-1)(x-4) ] = 1/6

=> ( -x+4+x-1 )/( x²-5x+4) = 1/6

=> 3/(x²-5x+4) = 1/6

=> 18 = x² - 5x + 4

=> x² - 5x + 4 - 18 = 0

=> x² - 5x - 14 = 0

Splitting the middle term , we get

=> x² - 7x + 2x - 14 = 0

=> x( x - 7 ) + 2( x - 7 ) = 0

=> ( x - 7 )( x + 2 ) = 0

=> x - 7 = 0 or x + 2 = 0

=> x = 7 or x = -2

Therefore ,

x = 7 or x = -2

••••