(√(4x+((12)/x)))=((x^2 +7)/(x+1)) x=?


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Question Number 64130 by ANTARES VY last updated on 14/Jul/19

$\sqrt{4 x + \frac{12}{x}} = \frac{x^{2} + 7}{x + 1}$ $x = ?$
Commented by Prithwish sen last updated on 14/Jul/19

$x = 1, 1, 3 and \frac{- 1 \pm 15 i}{2}$
	Answered by mr W last updated on 14/Jul/19

	$2 \sqrt{\frac{x^{2} + 3}{x}} = \frac{x^{2} + 7}{x + 1}$ $x = 1 a n d 3 a r e s o l u t i o n s .$ $\frac{4 (x^{2} + 3)}{x} = \frac{x^{4} + 14 x^{2} + 49}{x^{2} + 2 x + 1}$ $x^{5} + 14 x^{3} + 49 x = 4 x^{4} + 8 x^{3} + 4 x^{2} + 12 x^{2} + 24 x + 12$ $x^{5} - 4 x^{4} + 6 x^{3} - 16 x^{2} + 25 x - 12 = 0$ $. . . . . .$ ${(x - 1)}^{2} (x - 3) (x^{2} + x + 4) = 0$ $\Rightarrow x = 1, 3$ $x^{2} + x + 4 = 0$ $n o o t h e r r e a l s o l u t i o n .$
	Answered by ajfour last updated on 14/Jul/19

	$4 (x^{2} + 3) {(x + 1)}^{2} = x {(x^{2} + 7)}^{2}$ $l e t 4 a^{2} (x^{2} + 3) = x$ $& x + 1 = a (x^{2} + 7)$ $\Rightarrow x = \frac{1}{8 a^{2}} \pm \sqrt{\frac{1}{64 a^{4}} - 3}$ $x = \frac{1}{2 a} \pm \sqrt{\frac{1}{4 a^{2}} + \frac{1}{a} - 7}$ $l e t t h e i r r a t i o n a l & r a t i o n a l p a r t s b e e q u a l$ $\Rightarrow a = \frac{1}{4} & w e s e e t h a t f o r t h i s$ $v a l u e t h e i r r a t i o n a l p a r t s a r e$ $b o t h s a m e (= 1) .$ $h e n c e x = 2 \pm 1 .$ $\Rightarrow r e a l r o o t s a r e x = 1, 3 .$
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