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 36259 by behi83417@gmail.com last updated on 30/May/18

Answered by ajfour last updated on 30/May/18

let a+b=c+d=s ... (i) cd(ab−1)=ab ...(ii) ((ad+bc)/(bd))=((a+c)/(b+d)) ....(iii) ((a^2 +b^2 )/(ab))=((c^2 +d^2 )/(cd)) ....(iv) using (i) and (ii) in (iv): ((s^2 −2ab)/(ab))=((s^2 −((2ab)/(ab−1)))/((ab)/(ab−1))) ⇒ (s^2 /(ab))=((s^2 (ab−1))/(ab)) ⇒ ab=2 , hence cd=2 ...(v) from (iii): abd+b^2 c+bcd+ad^2 =abd+bcd ⇒ b^2 c+ad^2 =0 b^2 c+(8/(bc^2 ))=0 ⇒ bc=−2 using (v) in (i) : (a−b)^2 =(c−d)^2 =s^2 −8 or ((2/b)−b)^2 =(c−(2/c))^2 =s^2 −8 ⇒ b=c=±i(√2) a=d=∓i(√2) .

$${let}\:\:\:\:{a}+{b}={c}+{d}={s}\:\:\:\:\:\:\:\:\:\:\:...\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{cd}\left({ab}−\mathrm{1}\right)={ab}\:\:\:\:\:\:\:\:\:\:...\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{ad}+{bc}}{{bd}}=\frac{{a}+{c}}{{b}+{d}}\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({iii}\right) \\ $$$$\:\:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{ab}}=\frac{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }{{cd}}\:\:\:\:\:\:\:\:\:\:\:....\left({iv}\right) \\ $$$${using}\:\:\:\left({i}\right)\:{and}\:\left({ii}\right)\:{in}\:\left({iv}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{{s}^{\mathrm{2}} −\mathrm{2}{ab}}{{ab}}=\frac{{s}^{\mathrm{2}} −\frac{\mathrm{2}{ab}}{{ab}−\mathrm{1}}}{\frac{{ab}}{{ab}−\mathrm{1}}} \\ $$$$\Rightarrow\:\:\:\:\frac{{s}^{\mathrm{2}} }{{ab}}=\frac{{s}^{\mathrm{2}} \left({ab}−\mathrm{1}\right)}{{ab}} \\ $$$$\Rightarrow\:\:\:\:{ab}=\mathrm{2}\:\:\:,\:\:{hence}\:\:{cd}=\mathrm{2}\:\:\:\:...\left({v}\right) \\ $$$${from}\:\:\left({iii}\right): \\ $$$$\:\:{abd}+{b}^{\mathrm{2}} {c}+{bcd}+{ad}^{\mathrm{2}} ={abd}+{bcd} \\ $$$$\Rightarrow\:\:\:\:\:{b}^{\mathrm{2}} {c}+{ad}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} {c}+\frac{\mathrm{8}}{{bc}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:{bc}=−\mathrm{2} \\ $$$${using}\:\left({v}\right)\:\:{in}\:\:\left({i}\right)\:: \\ $$$$\:\:\:\:\left({a}−{b}\right)^{\mathrm{2}} =\left({c}−{d}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{8} \\ $$$${or}\:\:\:\:\left(\frac{\mathrm{2}}{{b}}−{b}\right)^{\mathrm{2}} =\left({c}−\frac{\mathrm{2}}{{c}}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} −\mathrm{8} \\ $$$$\Rightarrow\:\:{b}={c}=\pm{i}\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{a}={d}=\mp{i}\sqrt{\mathrm{2}}\:\:\:\:. \\ $$

Commented by ajfour last updated on 30/May/18

thanks sir, are the answers okay now ?

$${thanks}\:{sir},\:{are}\:{the}\:{answers}\:{okay} \\ $$$${now}\:? \\ $$

Commented by behi83417@gmail.com last updated on 30/May/18

alwyes the best.thanks in advance sir.

$${alwyes}\:{the}\:{best}.{thanks}\:{in}\:{advance}\:{sir}. \\ $$

Commented by behi83417@gmail.com last updated on 30/May/18

yes.it is smart and sweet method.

$${yes}.{it}\:{is}\:{smart}\:{and}\:{sweet}\:{method}. \\ $$

Commented by MJS last updated on 30/May/18

with your solution (3) is not defined because b+d=0

$$\mathrm{with}\:\mathrm{your}\:\mathrm{solution}\:\left(\mathrm{3}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{because} \\ $$$${b}+{d}=\mathrm{0} \\ $$

Commented by ajfour last updated on 30/May/18

a+c is even zero .

$${a}+{c}\:\:{is}\:{even}\:{zero}\:. \\ $$

Answered by MJS last updated on 30/May/18

(1) a=−b+c+d (2) ... (3) ... (4) b=c ∨ b=d case 1 b=c a=d (2) cd=2 ⇒ c=(2/d) (3) d^4 −2d^2 +4=0 ⇒ ⇒ d_1 =−((√6)/2)−((√2)/2)i; d_2 =−((√6)/2)+((√2)/2)i; d_3 =((√6)/2)−((√2)/2)i; d_4 =((√6)/2)+((√2)/2)i a_k =d_k b_k =c_k =(2/d_k )=conj(d_k ) case 2 b=d a=c (2) cd=2 ⇒ c=(2/d) (3) no solution

$$\left(\mathrm{1}\right)\:{a}=−{b}+{c}+{d} \\ $$$$\left(\mathrm{2}\right)\:... \\ $$$$\left(\mathrm{3}\right)\:... \\ $$$$\left(\mathrm{4}\right)\:{b}={c}\:\vee\:{b}={d} \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${b}={c} \\ $$$${a}={d} \\ $$$$\left(\mathrm{2}\right)\:{cd}=\mathrm{2}\:\Rightarrow\:{c}=\frac{\mathrm{2}}{{d}} \\ $$$$\left(\mathrm{3}\right)\:{d}^{\mathrm{4}} −\mathrm{2}{d}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{d}_{\mathrm{1}} =−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i};\:{d}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i}; \\ $$$${d}_{\mathrm{3}} =\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i};\:{d}_{\mathrm{4}} =\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{i} \\ $$$${a}_{{k}} ={d}_{{k}} \\ $$$${b}_{{k}} ={c}_{{k}} =\frac{\mathrm{2}}{{d}_{{k}} }=\mathrm{conj}\left({d}_{{k}} \right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${b}={d} \\ $$$${a}={c} \\ $$$$\left(\mathrm{2}\right)\:{cd}=\mathrm{2}\:\Rightarrow\:{c}=\frac{\mathrm{2}}{{d}} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{no}\:\mathrm{solution} \\ $$

Commented by behi83417@gmail.com last updated on 31/May/18

d^4 −2d^2 +4=0⇒d^4 −2d^2 +1=−3 (d^2 −1)^2 =−3⇒d^2 −1=±i(√3) d^2 =1±i(√3)⇒d=±(√(1±i(√3)))=^? ±((√6)/2)±((√2)/2)i

$${d}^{\mathrm{4}} −\mathrm{2}{d}^{\mathrm{2}} +\mathrm{4}=\mathrm{0}\Rightarrow{d}^{\mathrm{4}} −\mathrm{2}{d}^{\mathrm{2}} +\mathrm{1}=−\mathrm{3} \\ $$$$\left({d}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =−\mathrm{3}\Rightarrow{d}^{\mathrm{2}} −\mathrm{1}=\pm{i}\sqrt{\mathrm{3}} \\ $$$${d}^{\mathrm{2}} =\mathrm{1}\pm{i}\sqrt{\mathrm{3}}\Rightarrow{d}=\pm\sqrt{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}\overset{?} {=}\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{i} \\ $$

Answered by Rasheed.Sindhi last updated on 31/May/18

a≠0 ,b≠0,c≠0,d≠0 (∵ they′re denominators) (iii):(a/b)+(c/d)=((a+c)/(b+d))⇔(a/b)=(c/d) (a/b)=(c/d)=k (Say) (k≠0) a=bk,c=dk (i):a+b=c+d ⇒bk+b=dk+d ⇒b(k+1)=d(k+1) ⇒b=d ⇒a=c (ii) abcd=ab+cd ⇒bk.b.dk.d=bk.b+dk.d ⇒b^2 d^2 k^2 =(b^2 +d^2 )k ∵ d=b ⇒b^4 k^2 =2b^2 k ⇒b^4 k^2 −2b^2 k=0 ⇒b^2 k(b^2 −2)=0 ⇒b^2 =0 ∨ k=0 ∨ b^2 −2=0 ⇒b=0 ∨ k=0 ∨ b=±(√2) ∵ b≠0 ∨ k≠0 ⇒ b=±(√2) b=±(√2) d=±(√2) (iv):(a/b)+(b/a)=(c/d)+(d/c) ⇒((bk)/b)+(b/(bk))=(dk/d)+(d/dk) k+(1/k)=k+(1/k) Thus (a/b)+(c/d)=((a+c)/(b+d))⇒(a/b)+(b/a)=(c/d)+(d/c) (Always) (ii):abcd=ab+cd For a=c &b=d=±(√2) ⇒a^2 (±(√2))^2 =a(±(√2))+a(±(√2)) ⇒a^2 (±(√2))^2 =2a(±(√2)) ⇒a(±(√2))=2 ⇒a=±(2/(√2))×((√2)/(√2))=((±2(√2))/2)=±(√2) ⇒a=c=±(√2) Results: a= b=c=d=±(√2)

$${a}\neq\mathrm{0}\:,{b}\neq\mathrm{0},{c}\neq\mathrm{0},{d}\neq\mathrm{0}\:\left(\because\:\mathrm{they}'\mathrm{re}\:\mathrm{denominators}\right) \\ $$$$\left({iii}\right):\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}}\Leftrightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}} \\ $$$$\frac{{a}}{{b}}=\frac{{c}}{{d}}={k}\:\left({Say}\right)\:\:\left({k}\neq\mathrm{0}\right) \\ $$$${a}={bk},\mathrm{c}={dk} \\ $$$$\left(\mathrm{i}\right):{a}+{b}={c}+{d} \\ $$$$\:\:\:\:\Rightarrow{bk}+{b}={dk}+{d} \\ $$$$\:\:\:\:\Rightarrow{b}\left({k}+\mathrm{1}\right)={d}\left({k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\Rightarrow{b}={d} \\ $$$$\:\:\:\:\Rightarrow{a}={c} \\ $$$$\left(\mathrm{ii}\right)\:{abcd}={ab}+{cd} \\ $$$$\:\:\:\:\:\Rightarrow{bk}.{b}.{dk}.{d}={bk}.{b}+{dk}.{d} \\ $$$$\:\:\:\:\:\Rightarrow{b}^{\mathrm{2}} {d}^{\mathrm{2}} {k}^{\mathrm{2}} =\left({b}^{\mathrm{2}} +{d}^{\mathrm{2}} \right){k} \\ $$$$\:\:\:\:\because\:{d}={b} \\ $$$$\:\:\:\:\Rightarrow{b}^{\mathrm{4}} {k}^{\mathrm{2}} =\mathrm{2}{b}^{\mathrm{2}} {k} \\ $$$$\:\:\:\:\Rightarrow{b}^{\mathrm{4}} {k}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {k}=\mathrm{0} \\ $$$$\:\:\:\:\Rightarrow{b}^{\mathrm{2}} {k}\left({b}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{0} \\ $$$$\:\:\:\Rightarrow{b}^{\mathrm{2}} =\mathrm{0}\:\vee\:{k}=\mathrm{0}\:\vee\:{b}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\Rightarrow{b}=\mathrm{0}\:\vee\:{k}=\mathrm{0}\:\vee\:{b}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\because\:{b}\neq\mathrm{0}\:\vee\:{k}\neq\mathrm{0}\:\Rightarrow\:{b}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:{b}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:{d}=\pm\sqrt{\mathrm{2}} \\ $$$$\left(\mathrm{iv}\right):\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{{c}}{{d}}+\frac{{d}}{{c}} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\frac{{bk}}{{b}}+\frac{{b}}{{bk}}=\frac{{dk}}{{d}}+\frac{{d}}{{dk}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}+\frac{\mathrm{1}}{{k}}={k}+\frac{\mathrm{1}}{{k}} \\ $$$$\mathrm{Thus}\:\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}}\Rightarrow\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{{c}}{{d}}+\frac{{d}}{{c}}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{Always}\right) \\ $$$$\left(\mathrm{ii}\right):{abcd}={ab}+{cd} \\ $$$${F}\mathrm{or}\:{a}={c}\:\&{b}={d}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\Rightarrow{a}^{\mathrm{2}} \left(\pm\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={a}\left(\pm\sqrt{\mathrm{2}}\right)+{a}\left(\pm\sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow{a}^{\mathrm{2}} \left(\pm\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{a}\left(\pm\sqrt{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\Rightarrow{a}\left(\pm\sqrt{\mathrm{2}}\right)=\mathrm{2} \\ $$$$\:\:\:\:\:\:\Rightarrow{a}=\pm\frac{\mathrm{2}}{\sqrt{\mathrm{2}}}×\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}}=\frac{\pm\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\Rightarrow{a}={c}=\pm\sqrt{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${Results}:\:{a}=\:{b}={c}={d}=\pm\sqrt{\mathrm{2}} \\ $$

Commented by ajfour last updated on 31/May/18

(a/b)+(c/d)=((a+c)/(b+d)) ⇒ (a/b)=(c/d) how ? abd+ad^2 +bcd+b^2 c=abd+bcd ⇒ ad^2 =−b^2 c ⇒ (a/b)=−((bc)/d^2 ) if =(c/d) ⇒ ((−b)/d)=1 −b=d ...(1) also you found b=d ...(2) adding (1) & (2) 2d=0 ⇒ d=0 , b=0 not permitted by question!

$$\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}}\:\:\:\Rightarrow\:\:\frac{{a}}{{b}}=\frac{{c}}{{d}} \\ $$$${how}\:? \\ $$$${abd}+{ad}^{\mathrm{2}} +{bcd}+{b}^{\mathrm{2}} {c}={abd}+{bcd} \\ $$$$\Rightarrow\:\:\:{ad}^{\mathrm{2}} =−{b}^{\mathrm{2}} {c} \\ $$$$\Rightarrow\:\:\:\:\frac{{a}}{{b}}=−\frac{{bc}}{{d}^{\mathrm{2}} }\:\:\:\:{if}\:=\frac{{c}}{{d}} \\ $$$$\Rightarrow\:\:\frac{−{b}}{{d}}=\mathrm{1} \\ $$$$−{b}={d}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\left(\mathrm{1}\right) \\ $$$${also}\:{you}\:{found}\:\:{b}={d}\:\:\:\:\:\:\:...\left(\mathrm{2}\right) \\ $$$${adding}\:\:\left(\mathrm{1}\right)\:\&\:\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\mathrm{2}{d}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\:{d}=\mathrm{0}\:\:\:,\:\:{b}=\mathrm{0}\:\: \\ $$$${not}\:{permitted}\:{by}\:{question}! \\ $$

Commented by Rasheed.Sindhi last updated on 31/May/18

Mistake Sir! Only (a/b)=(c/d)⇒((a+c)/(b+d)) But (a/b)+(c/d)=((a+c)/(b+d))⇏(a/b)=(c/d) Thanks for correction!

$$\mathrm{Mistake}\:\mathrm{Sir}! \\ $$$$\mathrm{Only} \\ $$$$\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}+{c}}{{b}+\mathrm{d}} \\ $$$$\mathrm{But} \\ $$$$\:\:\:\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}}\nRightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}} \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{correction}! \\ $$

Commented by ajfour last updated on 31/May/18

(a/b)+(c/d)=((a+c)/(b+d)) if a=b=c=d then 1+1=1 .

$$\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}} \\ $$$${if}\:\:{a}={b}={c}={d}\:\:{then} \\ $$$$\mathrm{1}+\mathrm{1}=\mathrm{1}\:. \\ $$

Commented by Rasheed.Sindhi last updated on 31/May/18

You′re right!

$$\mathrm{You}'\mathrm{re}\:\mathrm{right}! \\ $$

Commented by MJS last updated on 31/May/18

because it′s a denominator too. usually it′s the first step to find the domain before any transformation. otherwise a, b, c, d =0 would also be ok

$$\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{denominator}\:\mathrm{too}.\:\mathrm{usually} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{first}\:\mathrm{step}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\boldsymbol{\mathrm{before}} \\ $$$$\mathrm{any}\:\mathrm{transformation}.\:\mathrm{otherwise}\:{a},\:{b},\:{c},\:{d}\:=\mathrm{0} \\ $$$$\mathrm{would}\:\mathrm{also}\:\mathrm{be}\:\mathrm{ok} \\ $$

Commented by MJS last updated on 31/May/18

a≠0 b≠0 c≠0 d≠0 b+d≠0

$${a}\neq\mathrm{0} \\ $$$${b}\neq\mathrm{0} \\ $$$${c}\neq\mathrm{0} \\ $$$${d}\neq\mathrm{0} \\ $$$${b}+{d}\neq\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 31/May/18

Sir, why b+d≠0 ?

$$\mathrm{Sir},\:\mathrm{why}\:{b}+{d}\neq\mathrm{0}\:? \\ $$

Commented by ajfour last updated on 31/May/18

as long as a+c=0 , b+d=0 is okay.

$${as}\:{long}\:{as}\:{a}+{c}=\mathrm{0}\:\:,\:\:{b}+{d}=\mathrm{0}\:\:{is}\:{okay}. \\ $$

Commented by MJS last updated on 31/May/18

(0/0) is not defined, so it′s not ok we can agree on b+d=0 is ok, but then a, b, c, d =0 should be allowed too whenever they happen to be denominators of a fraction with numerator=0 still, usually (p/q) is not defined when q=0, independent of the value of p f(x)=((x+1)/x) is not defined with x=0 ⇒ g(x)=(x/x) is not defined with x=0 but for some reasons it′s common to shorten it to g(x)=1

$$\frac{\mathrm{0}}{\mathrm{0}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined},\:\mathrm{so}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{ok} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{agree}\:\mathrm{on}\:{b}+{d}=\mathrm{0}\:\mathrm{is}\:\mathrm{ok},\:\mathrm{but}\:\mathrm{then} \\ $$$${a},\:{b},\:{c},\:{d}\:=\mathrm{0}\:\mathrm{should}\:\mathrm{be}\:\mathrm{allowed}\:\mathrm{too}\:\mathrm{whenever} \\ $$$$\mathrm{they}\:\mathrm{happen}\:\mathrm{to}\:\mathrm{be}\:\mathrm{denominators}\:\mathrm{of}\:\mathrm{a}\:\mathrm{fraction} \\ $$$$\mathrm{with}\:\mathrm{numerator}=\mathrm{0} \\ $$$$\mathrm{still},\:\mathrm{usually}\:\frac{{p}}{{q}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{when}\:{q}=\mathrm{0}, \\ $$$$\mathrm{independent}\:\mathrm{of}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{p} \\ $$$$ \\ $$$${f}\left({x}\right)=\frac{{x}+\mathrm{1}}{{x}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{with}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\:{g}\left({x}\right)=\frac{{x}}{{x}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{with}\:{x}=\mathrm{0} \\ $$$$\mathrm{but}\:\mathrm{for}\:\mathrm{some}\:\mathrm{reasons}\:\mathrm{it}'\mathrm{s}\:\mathrm{common}\:\mathrm{to} \\ $$$$\mathrm{shorten}\:\mathrm{it}\:\mathrm{to}\:{g}\left({x}\right)=\mathrm{1} \\ $$

Commented by ajfour last updated on 02/Jun/18

Prakash Sir, please throw some light in this matter, i m not entirely convinced. Isn′t my solution to this question acceptable?

$${Prakash}\:{Sir},\:{please}\:{throw}\:{some} \\ $$$${light}\:{in}\:{this}\:{matter},\:{i}\:{m}\:{not} \\ $$$${entirely}\:{convinced}.\:{Isn}'{t}\:{my} \\ $$$${solution}\:{to}\:{this}\:{question}\:{acceptable}? \\ $$

Answered by Rasheed.Sindhi last updated on 31/May/18

a+b=c+d.........(i) abcd=ab+cd......(ii) (a/b)+(c/d)=((a+c)/(b+d)).......(iii) (a/b)+(b/a)=(c/d)+(d/c)......(iv) ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗ (iv): (a/b)+(b/a)=(c/d)+(d/c) Let (a/b)=u & (c/d)=v u+(1/u)=v+(1/v) u^2 −(v+(1/v))u+1=0 u=((v+(1/v)±(√((v+(1/v))^2 −4(1)(1))))/(2(1))) u=((v+(1/v)±(√(v^2 +2+(1/v^2 )−4)))/2) u=((v+(1/v)±(√((v−(1/v))^2 )))/2) u=((v+(1/v)±(v−(1/v)))/2) u=((v+(1/v)±v∓(1/v)))/2)=v,(1/v) (a/b)=(c/d) ∣ (a/b)=(d/c) ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗ C-1: (a/b)=(c/d) (a/b)=(c/d)=k ⇒ a=bk & c=dk (i):a+b=c+d⇒bk+b=dk+d b(k+1)=d(k+1) b(k+1)−d(k+1)=0 (k+1)(b−d)=0 k=−1 ∨ b=d (a/b)=(c/d)=−1 ( a=−b ∧c=−d)∨ b=d a+b=c+d⇒c+d=0 ∨ a=c (a=−b ∧ c=−d) ∨ (a=c ∧ b=d) a=c ∧ b=d⇒ (ii):abcd=ab+cd⇒a^2 b^2 =2ab ⇒ab=2 (cd=2) (iii)(a/b)+(c/d)=((a+c)/(b+d)) ⇒((bk)/b)+(dk/d)=((bk+dk)/(b+d)) ⇒2k=k k=0 (a/b)=(c/d)=k=0 a=c=0 ∵a≠0,c≠0 a=−b ∧ c=−d (i) a.−a.c.−c=a.−a +c.−c a^2 c^2 =−a^2 −c^2 a^2 c^2 +a^2 =−c^2 ⇒a^2 =−(c^2 /(c^2 +1)) C-2: (a/b)=(d/c) Continue

$${a}+{b}={c}+{d}.........\left({i}\right) \\ $$$${abcd}={ab}+{cd}......\left({ii}\right) \\ $$$$\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}}.......\left({iii}\right) \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{{c}}{{d}}+\frac{{d}}{{c}}......\left({iv}\right) \\ $$$$\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast \\ $$$$\left({iv}\right):\:\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{{c}}{{d}}+\frac{{d}}{{c}} \\ $$$$\mathrm{Let}\:\frac{{a}}{{b}}={u}\:\:\&\:\:\frac{{c}}{{d}}={v} \\ $$$$\:\:\:\:\:\:{u}+\frac{\mathrm{1}}{{u}}={v}+\frac{\mathrm{1}}{{v}} \\ $$$$\:\:\:\:\:\:{u}^{\mathrm{2}} −\left({v}+\frac{\mathrm{1}}{{v}}\right){u}+\mathrm{1}=\mathrm{0} \\ $$$$\:{u}=\frac{{v}+\frac{\mathrm{1}}{{v}}\pm\sqrt{\left({v}+\frac{\mathrm{1}}{{v}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{1}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:{u}=\frac{{v}+\frac{\mathrm{1}}{{v}}\pm\sqrt{{v}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{v}^{\mathrm{2}} }−\mathrm{4}}}{\mathrm{2}} \\ $$$$\:{u}=\frac{{v}+\frac{\mathrm{1}}{{v}}\pm\sqrt{\left({v}−\frac{\mathrm{1}}{{v}}\right)^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\:{u}=\frac{{v}+\frac{\mathrm{1}}{{v}}\pm\left({v}−\frac{\mathrm{1}}{{v}}\right)}{\mathrm{2}} \\ $$$$\:{u}=\frac{\left.{v}+\frac{\mathrm{1}}{{v}}\pm{v}\mp\frac{\mathrm{1}}{{v}}\right)}{\mathrm{2}}={v},\frac{\mathrm{1}}{{v}} \\ $$$$\frac{{a}}{{b}}=\frac{{c}}{{d}}\:\:\:\:\:\:\:\mid\:\:\:\:\:\frac{{a}}{{b}}=\frac{{d}}{{c}} \\ $$$$\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast\ast \\ $$$${C}-\mathrm{1}:\:\frac{{a}}{{b}}=\frac{{c}}{{d}} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}={k}\:\Rightarrow\:{a}={bk}\:\&\:{c}={dk} \\ $$$$\left({i}\right):{a}+{b}={c}+{d}\Rightarrow{bk}+{b}={dk}+{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}\left({k}+\mathrm{1}\right)={d}\left({k}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{b}\left({k}+\mathrm{1}\right)−{d}\left({k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\left({k}+\mathrm{1}\right)\left({b}−{d}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{k}=−\mathrm{1}\:\vee\:{b}={d} \\ $$$$\:\:\:\:\:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}=−\mathrm{1}\: \\ $$$$\:\:\:\:\:\:\left(\:{a}=−{b}\:\wedge{c}=−{d}\right)\vee\:{b}={d} \\ $$$$\:\:\:\:\:\:\:{a}+{b}={c}+{d}\Rightarrow{c}+{d}=\mathrm{0}\:\vee\:{a}={c} \\ $$$$\:\:\:\:\:\:\:\left({a}=−{b}\:\wedge\:{c}=−{d}\right)\:\vee\:\:\left({a}={c}\:\wedge\:{b}={d}\right) \\ $$$$\:\:\:{a}={c}\:\wedge\:{b}={d}\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left({ii}\right):{abcd}={ab}+{cd}\Rightarrow{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{2}{ab} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{ab}=\mathrm{2}\:\left({cd}=\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left({iii}\right)\frac{{a}}{{b}}+\frac{{c}}{{d}}=\frac{{a}+{c}}{{b}+{d}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\frac{{bk}}{{b}}+\frac{{dk}}{{d}}=\frac{{bk}+{dk}}{{b}+{d}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}{k}={k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{a}}{{b}}=\frac{{c}}{{d}}={k}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}={c}=\mathrm{0}\:\because{a}\neq\mathrm{0},{c}\neq\mathrm{0} \\ $$$$\:\:\:{a}=−{b}\:\wedge\:{c}=−{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left({i}\right)\:{a}.−{a}.{c}.−{c}={a}.−{a}\:+{c}.−{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} {c}^{\mathrm{2}} =−{a}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} =−{c}^{\mathrm{2}} \Rightarrow{a}^{\mathrm{2}} =−\frac{{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${C}-\mathrm{2}:\:\frac{{a}}{{b}}=\frac{{d}}{{c}} \\ $$$${Continue} \\ $$

Commented by Rasheed.Sindhi last updated on 31/May/18

Please see that the approach used between the star lines(∗∗∗∗∗∗∗...) is correct ? That is (a/b)+(b/a)=(c/d)+(d/c)⇒(a/b)=(c/d) ?

$$\mathrm{Please}\:\mathrm{see}\:\mathrm{that}\:\mathrm{the}\:\mathrm{approach}\:\mathrm{used}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\:\mathrm{star}\:\mathrm{lines}\left(\ast\ast\ast\ast\ast\ast\ast...\right) \\ $$$$\mathrm{is}\:\mathrm{correct}\:?\:\mathrm{That}\:\mathrm{is} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\frac{{c}}{{d}}+\frac{{d}}{{c}}\Rightarrow\frac{{a}}{{b}}=\frac{{c}}{{d}}\:\:\:\:? \\ $$

Commented by behi83417@gmail.com last updated on 31/May/18

thanks for your work sir.yes it is true.

$${thanks}\:{for}\:{your}\:{work}\:{sir}.{yes}\:{it}\:{is}\:{true}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com