Given x≠y and x^2 =25x+y, y^2 =x+25y solve for the value of (√(x^2 +y^2 +1)) without using calculators or tools. Show your method.


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Question Number 138193 by mey3nipaba last updated on 10/Apr/21

$G i v e n x \neq y a n d x^{2} = 25 x + y, y^{2} = x + 25 y$ $s o l v e f o r t h e v a l u e o f \sqrt{x^{2} + y^{2} + 1} w i t h o u t$ $u s i n g c a l c u l a t o r s o r t o o l s .$ $S h o w y o u r m e t h o d .$
	Answered by liberty last updated on 11/Apr/21

	$x^{2} = 25 x + y$ $y^{2} = x + 25 y$ $\Rightarrow x^{2} + y^{2} = 26 (x + y)$ $\Rightarrow x^{2} - y^{2} = 24 (x - y)$ $\Rightarrow (x - y) [(x + y) - 24] = 0$ $\Rightarrow x + y = 24$ $\sqrt{x^{2} + y^{2} + 1} = \sqrt{26.24 + 1}$ $= \sqrt{(25 + 1) (25 - 1) + 1}$ $= \sqrt{25^{2}} = 25$
	Commented by mey3nipaba last updated on 11/Apr/21

	$t h a n k y o u v e r y m u c h . I a p p r e c i a t e .$
	Commented by mey3nipaba last updated on 11/Apr/21

	$B u t p l e a s e I d o n o t u n d e r s t a n d f r o m w h e r e$ $y o u e q u a t e d i t t o 0 . C a n y o u e x p l a i n ?$
	Commented by liberty last updated on 11/Apr/21

	$\Rightarrow x^{2} - y^{2} - 24 (x - y) = 0$ $\Rightarrow (x - y) (x + y) - 24 (x - y) = 0$ $\Rightarrow (x - y) (x + y - 24) = 0$ $s i n c e x \neq y i t f o l l o w s t h a t x + y - 24 = 0$
	Commented by mey3nipaba last updated on 11/Apr/21

	$T h a n k s . I g e t t h i s n o w b u t t h e p a r t w h e r e$ $y o u h a d 26.4 a s t h e v a l u e o f x^{2} + y^{2} .$ ${I s n}^{'} t i t s u p p o s e d t o b e 264 ?$
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