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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
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.
$${Sum}\:{of}\:{two}\:{irrational}\:{numbers}\:{is}\:\mathrm{1} \\ $$$${less}\:{than}\:{their}\:{product},\:{and}\:\mathrm{8}\:{less}\:{than} \\ $$$${their}\:{sum}\:{of}\:{squares}.\:{Find}\:{the}\:{larger} \\ $$$${of}\:{the}\:{two}\:{numbers}. \\ $$
Commented by nikif99 last updated on 27/Oct/23
a+b=ab−1 ⇒b=((a+1)/(a−1)) (1) ab−1=a^2 +b^2 −8 ⇒^((1)) a((a+1)/(a−1))=a^2 +(((a+1)/(a−1)))^2 −8 ⇒ ((a(a+1)(a−1))/((a−1)^2 ))−(((a−1)^2 )/((a−1)^2 ))=((a^2 (a−1)^2 )/((a−1)^2 ))+(((a+1)^2 )/((a−1)^2 ))−((8(a−1)^2 )/((a−1)^2 )) ⇒ a(a^2 −1)+(7−a^2 )(a−1)^2 −(a+1)^2 =0 ⇒ a^4 −3a^3 −5a^2 +17a−6=0 ⇒^(Peano) a=2∨a=3, ∈N rejected ((a^4 −3a^3 −5a^2 +17a−6)/((a−2)(a−3)))=a^2 +2a−1 ⇒a=−1+(√2)∨a=−1−(√2) greatest a=−1+(√2)
$${a}+{b}={ab}−\mathrm{1}\:\Rightarrow{b}=\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}\:\left(\mathrm{1}\right) \\ $$$${ab}−\mathrm{1}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{8}\:\overset{\left(\mathrm{1}\right)} {\Rightarrow}{a}\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}={a}^{\mathrm{2}} +\left(\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{8}\:\Rightarrow \\ $$$$\frac{{a}\left({a}+\mathrm{1}\right)\left({a}−\mathrm{1}\right)}{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{{a}^{\mathrm{2}} \left({a}−\mathrm{1}\right)^{\mathrm{2}} }{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }−\frac{\mathrm{8}\left({a}−\mathrm{1}\right)^{\mathrm{2}} }{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$${a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)+\left(\mathrm{7}−{a}^{\mathrm{2}} \right)\left({a}−\mathrm{1}\right)^{\mathrm{2}} −\left({a}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\:\Rightarrow \\ $$$${a}^{\mathrm{4}} −\mathrm{3}{a}^{\mathrm{3}} −\mathrm{5}{a}^{\mathrm{2}} +\mathrm{17}{a}−\mathrm{6}=\mathrm{0}\:\overset{{Peano}} {\Rightarrow}\:{a}=\mathrm{2}\vee{a}=\mathrm{3},\:\in\mathbb{N}\:{rejected} \\ $$$$\frac{{a}^{\mathrm{4}} −\mathrm{3}{a}^{\mathrm{3}} −\mathrm{5}{a}^{\mathrm{2}} +\mathrm{17}{a}−\mathrm{6}}{\left({a}−\mathrm{2}\right)\left({a}−\mathrm{3}\right)}={a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}\:\Rightarrow{a}=−\mathrm{1}+\sqrt{\mathrm{2}}\vee{a}=−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$${greatest}\:{a}=−\mathrm{1}+\sqrt{\mathrm{2}} \\ $$
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)
$${a},{b}\:{are}\:{two}\:{irrational}\:{numbers} \\ $$$${a}+{b}={ab}−\mathrm{1} \\ $$$${a}+{b}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{8} \\ $$$${a}+{b}=\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}{ab}−\mathrm{8} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} −\left({a}+{b}\right)−\mathrm{2}{ab}−\mathrm{8}=\mathrm{0} \\ $$$${Let} \\ $$$${a}+{b}={t}={ab}−\mathrm{1}\Rightarrow{ab}={t}+\mathrm{1} \\ $$$${t}^{\mathrm{2}} −{t}−\mathrm{2}\left({t}+\mathrm{1}\right)−\mathrm{8}=\mathrm{0} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{10}=\mathrm{0} \\ $$$$\left({t}−\mathrm{5}\right)\left({t}+\mathrm{2}\right)=\mathrm{0} \\ $$$${t}={a}+{b}={ab}−\mathrm{1}=\mathrm{5}\:,\:−\mathrm{2} \\ $$$$\begin{cases}{{a}+{b}=\mathrm{5}\Rightarrow{ab}=\mathrm{6}\Rightarrow\left({a},{b}\right)=\left(\mathrm{2},\mathrm{3}\right)\:\mathrm{or}\:\left(\mathrm{3},\mathrm{2}\right)^{\bigstar} }\\{{a}+{b}=−\mathrm{2}\Rightarrow{ab}=−\mathrm{1}^{\bigstar\bigstar} }\end{cases} \\ $$$$\:^{\bigstar} {a}+{b}=\mathrm{5}\:,\:{ab}=\mathrm{6}\:,\:\:{specify} \\ $$$$\:{integer}\:{values}\:{for}\:{a},{b} \\ $$$$\:{Hence}\:{rejected}. \\ $$$$\: \\ $$$$\:^{\bigstar\bigstar} \:\:{a}+{b}=−\mathrm{2},{ab}=−\mathrm{1}: \\ $$$${b}=−\frac{\mathrm{1}}{{a}} \\ $$$${a}−\frac{\mathrm{1}}{{a}}=−\mathrm{2} \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{1}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}}}{\mathrm{2}}=−\mathrm{1}\pm\sqrt{\mathrm{2}}\: \\ $$$${b}=−\frac{\mathrm{1}}{{a}}=−\frac{\mathrm{1}}{−\mathrm{1}\pm\sqrt{\mathrm{2}}}×\frac{−\mathrm{1}\mp\sqrt{\mathrm{2}}}{−\mathrm{1}\mp\sqrt{\mathrm{2}}}=−\frac{−\mathrm{1}\mp\sqrt{\mathrm{2}}}{\mathrm{1}−\mathrm{2}}=−\mathrm{1}\mp\sqrt{\mathrm{2}}\: \\ $$$$\left({a},{b}\right)=\left(−\mathrm{1}+\sqrt{\mathrm{2}\:}\:,\:−\mathrm{1}−\sqrt{\mathrm{2}}\:\right),\left(−\mathrm{1}−\sqrt{\mathrm{2}\:}\:,\:−\mathrm{1}+\sqrt{\mathrm{2}}\:\right) \\ $$$${Clearly}, \\ $$$${Larger}\:{value}=−\mathrm{1}+\sqrt{\mathrm{2}}\: \\ $$
Commented by Rasheed.Sindhi last updated on 27/Oct/23
necx sir please go through my solution once more. I′ve corrected it now!
$${necx}\:{sir}\:{please}\:{go}\:{through}\:{my} \\ $$$${solution}\:{once}\:{more}.\:{I}'{ve}\:{corrected} \\ $$$${it}\:{now}! \\ $$
Commented by necx122 last updated on 27/Oct/23
how is that possible sir? by merely looking at both values it seems like 1+(√2) is greater
$${how}\:{is}\:{that}\:{possible}\:{sir}?\:{by}\:{merely} \\ $$$${looking}\:{at}\:{both}\:{values}\:{it}\:{seems}\:{like} \\ $$$$\mathrm{1}+\sqrt{\mathrm{2}}\:{is}\:{greater} \\ $$
Commented by mr W last updated on 27/Oct/23
but the two numbers are −1±(√2). so −1+(√2) is the larger one and −1−(√2) is the smaller one.
$${but}\:{the}\:{two}\:{numbers}\:{are}\:−\mathrm{1}\pm\sqrt{\mathrm{2}}.\:{so} \\ $$$$−\mathrm{1}+\sqrt{\mathrm{2}}\:{is}\:{the}\:{larger}\:{one}\:{and} \\ $$$$−\mathrm{1}−\sqrt{\mathrm{2}}\:{is}\:{the}\:{smaller}\:{one}. \\ $$
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
You′re right mr W sir. I′ve found my mistake and now corrected it!
$${You}'{re}\:{right}\:\boldsymbol{{mr}}\:\boldsymbol{{W}}\:\boldsymbol{{sir}}.\:{I}'{ve}\:{found}\:{my}\: \\ $$$${mistake}\:{and}\:{now}\:{corrected}\:{it}! \\ $$

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