Question

Solve by 'completion of square' method:

$2. \: \frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}$

Standard:- 10

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Answer 1

Your answer is --

we have ,

1/(x+1) + 2(x+2) = 4/(x+4)

=> 1/(x+1) + 2(x+2) - 4/(x+4) = 0

=>(x+2)(x+4) + 2(x+1)(x+4) - 4(x+1)(x+2) = 0

=> x^2+4x+2x+8+2x^2+8x+2x+8 - 4x^2 -8x - 4x - 8 = 0

=> - x^2 + 4x + 8 = 0

=> x^2 - 4x - 8 = 0

Now, factorise this ☝ by completing square

=> x^2 - 4x = 8

=> x^2 - 2× 4x/2 = 8

=> x^2 - 2 × 4x/2 + (4/2)^2 = (4/2)^2 + 8

=> x^2 - 2 × 4x/2 + 4/2^2 = 2^2 + 8

{ °•° a^2 - 2ab + b^2 = (a-b)^2 }

=> (x - 4/2)^2 = 4 + 8

=> (x - 2)^2 = 12

=> x - 2 = ±√12

=> x = 2 ± √12 = 2 ± 2√3

Hence, x = 2+ 2√3 or 2 - 2√3

【 Hope it helps you 】