Solve: x^2 =a+(y−z)^2 y^2 =b+(z−x)^2 z^2 =c+(x−y)^2


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Question Number 215356 by Abdullahrussell last updated on 04/Jan/25

$S o l v e :$ $x^{2} = a + {(y - z)}^{2}$ $y^{2} = b + {(z - x)}^{2}$ $z^{2} = c + {(x - y)}^{2}$
	Answered by mr W last updated on 04/Jan/25

	$(x + y - z) (x - y + z) = a . . . (i)$ $(- x + y + z) (x + y - z) = b . . . (i i)$ $(x - y + z) (- x + y + z) = c . . . (i i i)$ $\Rightarrow {(- x + y + z)}^{2} {(x - y + z)}^{2} {(x + y - z)}^{2} = a b c$ $a s s u m e a b c ⩾ 0, o t h e r w i s e n o s o l u t i o n .$ $(- x + y + z) (x - y + z) (x + y - z) = \pm \sqrt{a b c}$ $\Rightarrow - x + y + z = \pm \frac{\sqrt{a b c}}{a} . . . (I)$ $\Rightarrow x - y + z = \pm \frac{\sqrt{a b c}}{b} . . . (I I)$ $\Rightarrow x + y - z = \pm \frac{\sqrt{a b c}}{c} . . . (I I I)$ $(I I) + (I I I) :$ $\Rightarrow 2 x = \pm (\frac{1}{b} + \frac{1}{c}) \sqrt{a b c}$ $x = \pm \frac{1}{2} (\frac{1}{b} + \frac{1}{c}) \sqrt{a b c}$ $s i m i l a r l y$ $y = \pm \frac{1}{2} (\frac{1}{c} + \frac{1}{a}) \sqrt{a b c}$ $z = \pm \frac{1}{2} (\frac{1}{a} + \frac{1}{b}) \sqrt{a b c}$
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