Sum of two irrational numbers is 1 less than their product, and 8 less than their sum of squares. Find the larger of the two numbers.


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Question Number 199011 by necx122 last updated on 26/Oct/23

$S u m o f t w o i r r a t i o n a l n u m b e r s i s 1$ $l e s s t h a n t h e i r p r o d u c t, a n d 8 l e s s t h a n$ $t h e i r s u m o f s q u a r e s . F i n d t h e l a r g e r$ $o f t h e t w o n u m b e r s .$
Commented by nikif99 last updated on 27/Oct/23

$a + b = a b - 1 \Rightarrow b = \frac{a + 1}{a - 1} (1)$ $a b - 1 = a^{2} + b^{2} - 8 \overset{(1)}{\Rightarrow} a \frac{a + 1}{a - 1} = a^{2} + {(\frac{a + 1}{a - 1})}^{2} - 8 \Rightarrow$ $\frac{a (a + 1) (a - 1)}{{(a - 1)}^{2}} - \frac{{(a - 1)}^{2}}{{(a - 1)}^{2}} = \frac{a^{2} {(a - 1)}^{2}}{{(a - 1)}^{2}} + \frac{{(a + 1)}^{2}}{{(a - 1)}^{2}} - \frac{8 {(a - 1)}^{2}}{{(a - 1)}^{2}} \Rightarrow$ $a (a^{2} - 1) + (7 - a^{2}) {(a - 1)}^{2} - {(a + 1)}^{2} = 0 \Rightarrow$ $a^{4} - 3 a^{3} - 5 a^{2} + 17 a - 6 = 0 \overset{P e a n o}{\Rightarrow} a = 2 \lor a = 3, \in N r e j e c t e d$ $\frac{a^{4} - 3 a^{3} - 5 a^{2} + 17 a - 6}{(a - 2) (a - 3)} = a^{2} + 2 a - 1 \Rightarrow a = - 1 + \sqrt{2} \lor a = - 1 - \sqrt{2}$ $g r e a t e s t a = - 1 + \sqrt{2}$
	Answered by Rasheed.Sindhi last updated on 27/Oct/23
	$a,b are two irrational numbers a+b=ab−1 a+b=a^2 +b^2 −8 a+b=(a+b)^2 −2ab−8 (a+b)^2 −(a+b)−2ab−8=0 Let a+b=t=ab−1⇒ab=t+1 t^2 −t−2(t+1)−8=0 t^2 −3t−10=0 (t−5)(t+2)=0 t=a+b=ab−1=5 , −2 { ((a+b=5⇒ab=6⇒(a,b)=(2,3) or (3,2)^★ )),((a+b=−2⇒ab=−1^(★★) )) :} ^★ a+b=5 , ab=6 , specify integer values for a,b Hence rejected. ^(★★) a+b=−2,ab=−1: b=−(1/a) a−(1/a)=−2 a^2 +2a−1=0 a=((−2±(√(4+4)))/2)=−1±(√2) b=−(1/a)=−(1/(−1±(√2)))×((−1∓(√2))/(−1∓(√2)))=−((−1∓(√2))/(1−2))=−1∓(√2) (a,b)=(−1+(√(2 )) , −1−(√2) ),(−1−(√(2 )) , −1+(√2) ) Clearly, Larger value=−1+(√2)$
	$a, b a r e t w o i r r a t i o n a l n u m b e r s$ $a + b = a b - 1$ $a + b = a^{2} + b^{2} - 8$ $a + b = {(a + b)}^{2} - 2 a b - 8$ ${(a + b)}^{2} - (a + b) - 2 a b - 8 = 0$ $L e t$ $a + b = t = a b - 1 \Rightarrow a b = t + 1$ $t^{2} - t - 2 (t + 1) - 8 = 0$ $t^{2} - 3 t - 10 = 0$ $(t - 5) (t + 2) = 0$ $t = a + b = a b - 1 = 5, - 2$ ${\begin{cases} a + b = 5 \Rightarrow a b = 6 \Rightarrow (a, b) = (2, 3) or {(3, 2)}^{★} \\ a + b = - 2 \Rightarrow a b = - 1^{★ ★} \end{cases}$ $^{★} a + b = 5, a b = 6, s p e c i f y$ $i n t e g e r v a l u e s f o r a, b$ $H e n c e r e j e c t e d .$ $^{★ ★} a + b = - 2, a b = - 1 :$ $b = - \frac{1}{a}$ $a - \frac{1}{a} = - 2$ $a^{2} + 2 a - 1 = 0$ $a = \frac{- 2 \pm \sqrt{4 + 4}}{2} = - 1 \pm \sqrt{2}$ $b = - \frac{1}{a} = - \frac{1}{- 1 \pm \sqrt{2}} \times \frac{- 1 \mp \sqrt{2}}{- 1 \mp \sqrt{2}} = - \frac{- 1 \mp \sqrt{2}}{1 - 2} = - 1 \mp \sqrt{2}$ $(a, b) = (- 1 + \sqrt{2}, - 1 - \sqrt{2}), (- 1 - \sqrt{2}, - 1 + \sqrt{2})$ $C l e a r l y,$ $L a r g e r v a l u e = - 1 + \sqrt{2}$
	Commented by Rasheed.Sindhi last updated on 27/Oct/23

	$n e c x s i r p l e a s e g o t h r o u g h m y$ $s o l u t i o n o n c e m o r e . I^{'} v e c o r r e c t e d$ $i t n o w!$
	Commented by necx122 last updated on 27/Oct/23

	$h o w i s t h a t p o s s i b l e s i r ? b y m e r e l y$ $l o o k i n g a t b o t h v a l u e s i t s e e m s l i k e$ $1 + \sqrt{2} i s g r e a t e r$
	Commented by mr W last updated on 27/Oct/23

	$b u t t h e t w o n u m b e r s a r e - 1 \pm \sqrt{2} . s o$ $- 1 + \sqrt{2} i s t h e l a r g e r o n e a n d$ $- 1 - \sqrt{2} i s t h e s m a l l e r o n e .$
	Commented by necx122 last updated on 27/Oct/23
	Yes, it's clear now. I think the error was just an oversight. Thank you, sir.
	Commented by Rasheed.Sindhi last updated on 27/Oct/23

	${Y o u}^{'} r e r i g h t m r W s i r . I^{'} v e f o u n d m y$ $m i s t a k e a n d n o w c o r r e c t e d i t!$
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