Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 205092 by BaliramKumar last updated on 08/Mar/24

	Answered by A5T last updated on 08/Mar/24

	$n^{2} + 19 n + 92 - x^{2} = 0$ $n = \frac{- 19 \underline{+} \sqrt{361 - 4 (92 - x^{2})}}{2} = \frac{- 19 \underline{+} \sqrt{4 x^{2} - 7}}{2}$ $4 x^{2} - 7 = p^{2} \Rightarrow (2 x - p) (2 x + p) = 7 = 1 \times 7 = - 1 \times - 7$ $2 x - p = 1 \land 2 x + p = 7 \Rightarrow x = 2$ $2 x - p = - 1 \land 2 x + p = - 7 \Rightarrow x = - 2$ $n = \frac{- 19 \underline{+} 3}{2} = - 8 o r - 11 \Rightarrow s u m = - 19 \Rightarrow (d)$
	Commented by BaliramKumar last updated on 08/Mar/24

	$Thanks sir$
	Answered by A5T last updated on 08/Mar/24

	${(n + 9)}^{2} < n^{2} + 19 n + 92 < {(n + 10)}^{2}$ $w h e n n ⩾ 0, s o n^{2} + 19 n + 92 {c a n}^{'} t b e a p e r f e c t s q u a r e$ ${(n + 10)}^{2} = n^{2} + 20 n + 100 = n^{2} + 19 n + 92 + n + 8$ $w h e n n = - 8, t h e n n^{2} + 19 n + 92 = {(- 8 + 10)}^{2} = 4$ $b u t w h e n n < - 8, {(n + 10)}^{2} < n^{2} + 19 n + 92$ ${(n + 9)}^{2} = n^{2} + 18 n + 81 = n^{2} + 19 n + 92 - n - 11$ $w h e n n = - 11, n^{2} + 19 n + 92 = {(- 11 + 9)}^{2} = 4$ $b u t w h e n n < - 11; {(n + 9)}^{2} > n^{2} + 19 n + 92$ $\Rightarrow w h e n n < - 11; {(n + 10)}^{2} < n^{2} + 19 n + 92 < {(n + 9)}^{2}$ $E q u a l i t y h o l d s a t e i t h e r - 8 o r - 11$ $S o, o n e c a n e a s i l y c h e c k n = - 10, - 9, - 7, - 6, . . . - 1$ $t o s e e t h a t n o n e s a t i s f i e s .$
	Commented by BaliramKumar last updated on 08/Mar/24

	$difficult method$ $example n^{2} + 2 n + 361 twelve value of n$
	Commented by A5T last updated on 08/Mar/24

	$T h i s s o r t o f m e t h o d c o u l d b e u s e f u l f o r q u e s t i o n s$ $l i k e n^{4} + n^{3} + n^{2} + n + 1 = x^{2} (Q 204397) w h e r e i t$ $w o u l d b e d i f f i c u l t t o m a k e n t h e s u b j e c t o f$ $f o r m u l a i n t e r m s o f x .$ $S o, m e t h o d s h a v e t h e i r d i s a d v a n t a g e s a n d$ $a d v a n t a g e s .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum