solve x+(√(x(x+1)))+(√(x(x+2)))+(√((x+1)(x+2)))=2


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Question Number 137415 by mr W last updated on 02/Apr/21

$s o l v e$ $x + \sqrt{x (x + 1)} + \sqrt{x (x + 2)} + \sqrt{(x + 1) (x + 2)} = 2$
Commented by MJS_new last updated on 03/Apr/21

$x_{1} = \frac{1}{24}$ $x_{2} \approx - 2.73908312241$ $x_{2} is a solution of x^{4} + x^{3} - 4 x^{2} + 2 x - \frac{1}{4} = 0$ $which I {couldn}^{'} t solve exactly and there are$ $no complex solutions .$
Commented by liberty last updated on 03/Apr/21

$h o w t o g e t x^{4} + x^{3} - 4 x^{2} + 2 x - \frac{1}{4} = 0 ?$
Commented by MJS_new last updated on 03/Apr/21

$\sqrt{A} + \sqrt{B} = D - \sqrt{C}$ $squaring, transforming$ $2 (\sqrt{A B} + D \sqrt{C}) = C + D^{2} - A - B$ $squaring, transforming$ $8 D \sqrt{A} \sqrt{B} \sqrt{C} = A^{2} + B^{2} + C^{2} + D^{4} - 2 (A B + A C + {A D}^{2} + B C + {B D}^{2} + {C D}^{2})$ $squaring, transforming, inserting$ $x^{5} + \frac{23}{24} x^{4} - \frac{97}{24} x^{3} + \frac{13}{6} x^{2} - \frac{1}{3} x + \frac{1}{96} = 0$ $trying factors of \frac{1}{96}$ $(x - \frac{1}{24}) (x^{4} + x^{3} - 4 x^{2} + 2 x - \frac{1}{4}) = 0$ $we can find 2 square factors$ $let x = z - \frac{1}{4}$ $z^{4} - \frac{35}{8} z^{2} + \frac{33}{8} z - \frac{259}{256} = 0$ $(z^{2} - a z - b) (z^{2} + a z - c) = 0$ $\Rightarrow$ $(1) a^{2} + b + c - \frac{35}{8} = 0$ $(2) a b - a c + \frac{33}{8} = 0$ $(3) b c + \frac{259}{256} = 0$ $(1) c = - a^{2} - b + \frac{35}{8}$ $(2) b = - \frac{a^{2}}{2} - \frac{33}{16 a} + \frac{35}{16} \Rightarrow c = - \frac{a^{2}}{2} + \frac{33}{16 a} + \frac{35}{16}$ $(3) \Rightarrow a^{6} - \frac{35}{4} a^{4} + \frac{371}{16} a^{2} - \frac{1089}{64} = 0$ $let a = \sqrt{r + \frac{35}{12}}$ $r^{3} - \frac{7}{3} r + \frac{107}{108} = 0$ $sadly this has no useable solution . . .$ $r_{1} = \frac{2 \sqrt{7}}{3} \sin \frac{\arcsin \frac{107 \sqrt{7}}{392}}{3}$ $r_{2} = \frac{2 \sqrt{7}}{3} \cos \frac{π + 2 \arcsin \frac{107 \sqrt{7}}{392}}{6}$ $r_{3} = - \frac{2 \sqrt{7}}{3} \sin \frac{π + \arcsin \frac{107 \sqrt{7}}{392}}{3}$ $\Rightarrow this makes no sense, {it}^{'} s ‘ ‘ {cheaper}^{″} to$ $approximately solve$ $x^{4} + x^{3} - 4 x^{2} + 2 x - \frac{1}{4} = 0$ $and check the solutions (squaring causes$ $false solutions)$ $x_{1} \approx - 2.73908 ★$ $x_{2} \approx . 200728 w r o n g$ $x_{3} \approx . 399135 w r o n g$ $x_{4} \approx 1.13922 w r o n g$
Commented by mr W last updated on 03/Apr/21

$t h a n k s s i r s!$
	Answered by mindispower last updated on 02/Apr/21

	$l e t Z = \sqrt{x (x + 1)} + \sqrt{x (x + 2)}$ $Y = x + \sqrt{(x + 1) (x + 2)}$ $Z^{2} = 2 x^{2} + 3 x + 2 x \sqrt{(x + 1) (x + 2)}$ $Y^{2} = 2 x^{2} + 3 x + 2 + 2 x \sqrt{(x + 2) (x + 1)}$ $Y^{2} - Z^{2} = 2 . . . 1$ $Y + Z = 2 . . . 2$ $(1) & (2) \Rightarrow Y - Z = 1$ $\Rightarrow Y = \frac{3}{2}$ $\Leftrightarrow (\frac{3}{2} - x) = \sqrt{(x + 1) (x + 2)} \Rightarrow - 3 x + \frac{9}{4} = 3 x + 2$ $\Rightarrow 6 x = \frac{1}{4} \Rightarrow x = \frac{1}{24} . . . .$ $\frac{1}{24} . . . i w i l l p o s t c o m p l e t s o l u t i o n l a t e r$ $i s o l v e d w i t h e \Rightarrow n o t \Leftrightarrow m u s t t c h e k i f \frac{1}{24} i s s o l u t i o n$
	Commented by mr W last updated on 02/Apr/21

	$t h a n k s s i r!$ $\frac{1}{24} i s a s o l u t o n . b u t i s i t t h e o n l y o n e ?$
	Answered by liberty last updated on 03/Apr/21

	$f o r x ⩾ 0$ $\sqrt{x} (\sqrt{x} + \sqrt{x + 1}) + \sqrt{x + 2} (\sqrt{x} + \sqrt{x + 1}) = 2$ $\sqrt{x} + \sqrt{x + 2} = \frac{2}{\sqrt{x} + \sqrt{x + 1}}$ $t h a t c a n b e w r i t t e n a s$ $\sqrt{x} + \sqrt{x + 2} = 2 \sqrt{x + 1} - 2 \sqrt{x}$ $3 \sqrt{x} + \sqrt{x + 2} = 2 \sqrt{x + 1}$ $10 x + 2 + 6 \sqrt{x (x + 2)} = 4 x + 4$ $3 \sqrt{x (x + 2)} = 1 - 3 x$ $9 x^{2} + 18 x = 9 x^{2} - 6 x + 1$ $\Leftrightarrow 24 x = 1; x = \frac{1}{24}$
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