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

Answered by ajfour last updated on 30/May/18

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

Commented by ajfour last updated on 30/May/18

$$thankssir,aretheanswersokay$$$$now?$$

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

$$alwyesthebest.thanksinadvancesir.$$

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

$$yes.itissmartandsweetmethod.$$

Commented by MJS last updated on 30/May/18

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

Commented by ajfour last updated on 30/May/18

$$a+cisevenzero.$$

Answered by MJS last updated on 30/May/18

$$\left(1\right)a=-b+c+d$$$$\left(2\right)...$$$$\left(3\right)...$$$$\left(4\right)b=c\vee b=d$$$$\mathrm{case}1$$$$b=c$$$$a=d$$$$\left(2\right)cd=2\Rightarrow c=\frac{2}{d}$$$$\left(3\right)d^4-2d^2+4=0\Rightarrow$$$$\Rightarrow d_1=-\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\mathrm{i};d_2=-\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\mathrm{i};$$$$d_3=\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}\mathrm{i};d_4=\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}\mathrm{i}$$$$a_k=d_k$$$$b_k=c_k=\frac{2}{d_k}=\mathrm{conj}\left(d_k\right)$$$$$$$$\mathrm{case}2$$$$b=d$$$$a=c$$$$\left(2\right)cd=2\Rightarrow c=\frac{2}{d}$$$$\left(3\right)\mathrm{no}\mathrm{solution}$$

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

$$d^4-2d^2+4=0\Rightarrow d^4-2d^2+1=-3$$$${(d^2-1)}^2=-3\Rightarrow d^2-1=\pm i\sqrt{3}$$$$d^2=1\pm i\sqrt{3}\Rightarrow d=\pm \sqrt{1\pm i\sqrt{3}}\stackrel{?}{=}\pm \frac{\sqrt{6}}{2}\pm \frac{\sqrt{2}}{2}i$$

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

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

Commented by ajfour last updated on 31/May/18

$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}\Rightarrow \frac{a}{b}=\frac{c}{d}$$$$how?$$$$abd+{ad}^2+bcd+b^{2}c=abd+bcd$$$$\Rightarrow {ad}^2=-b^{2}c$$$$\Rightarrow \frac{a}{b}=-\frac{bc}{d^2}if=\frac{c}{d}$$$$\Rightarrow \frac{-b}{d}=1$$$$-b=d...\left(1\right)$$$$alsoyoufoundb=d...\left(2\right)$$$$adding\left(1\right)\mathrm{\&}\left(2\right)$$$$2d=0\Rightarrow d=0,b=0$$$$notpermittedbyquestion!$$

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

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

Commented by ajfour last updated on 31/May/18

$$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$$$ifa=b=c=dthen$$$$1+1=1.$$

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

$${\mathrm{You}}^{\prime }\mathrm{re}\mathrm{right}!$$

Commented by MJS last updated on 31/May/18

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

Commented by MJS last updated on 31/May/18

$$a\ne 0$$$$b\ne 0$$$$c\ne 0$$$$d\ne 0$$$$b+d\ne 0$$

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

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

Commented by ajfour last updated on 31/May/18

$$aslongasa+c=0,b+d=0isokay.$$

Commented by MJS last updated on 31/May/18

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

Commented by ajfour last updated on 02/Jun/18

$$PrakashSir,pleasethrowsome$$$$lightinthismatter,imnot$$$$entirelyconvinced.{Isn}^{\prime }tmy$$$$solutiontothisquestionacceptable?$$

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

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

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

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

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

$$thanksforyourworksir.yesitistrue.$$

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