(1/((1−x)^2 ))−(4/((1+x)^2 ))=1


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Question Number 104621 by ajfour last updated on 22/Jul/20

$\frac{1}{{(1 - x)}^{2}} - \frac{4}{{(1 + x)}^{2}} = 1$
	Answered by prakash jain last updated on 22/Jul/20

	${(1 + x)}^{2} - 4 {(1 - x)}^{2} = (1 - 2 x + x^{2}) (1 + 2 x + x^{2})$ $- 3 x^{2} + 10 x - 3 =$ $x^{4} - 2 x^{2} + 1$ $x^{4} + x^{2} - 10 x + 4 = 0$ ${(x^{2} + 2)}^{2} - 3 x^{2} - 10 x = 0$ ${(x^{2} + 2)}^{2} + 2 k (x^{2} + 2) + k^{2}$ $= (3 + 2 k) x^{2} + 10 x + (k^{2} + 4 k)$ ${(x^{2} + 2 + k)}^{2} = (3 + 2 k) x^{2} + 10 x + (k^{2} + 4 k)$ $10 = 2 \sqrt{(3 + 2 k) (k^{2} + 4 k)} \Rightarrow k = 1$ ${(x^{2} + 3)}^{2} = 5 {(x + 1)}^{2}$ $Equation can be splitted into$ $2 quadratic at this point .$
	Commented by ajfour last updated on 24/Jul/20

	$t h a n k s! p r a k a s h S i r, b u t g e n e r a l l y$ $f i n d i n g k, w o u l d r e q u i r e, s o l v i n g a$ $c u b i c, i b e l i e v e . .$
	Commented by prakash jain last updated on 24/Jul/20

	$I found as solution was obvious .$ $You are right .$
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