Question Number 49147 by behi83417@gmail.com last updated on 03/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Dec/18
![ab−ad+cd−bc=0 a(b−d)−c(b−d)=0 (a−c)(b−d)=0 either a=c or b=d let a=c a^2 +d^2 =c^2 +d^2 so b=d henc a=c b=d a+b+c+d=1 2a+2b=1 a+b=(1/2) ac+bd=1 a^2 +b^2 =1 a^2 +((1/2)−a)^2 =1 [a+b=(1/2)] a^2 +(1/4)−a+a^2 =1 8a^2 −4a+1=4 8a^2 −4a−3=0 a=((4±(√(16+96)))/(16))=((4±(√(112)))/(16))=((4±4(√7))/(16)) a=((1±(√7))/4) c=((1±(√7))/4) b=(1/2)−(a) b=(1/2)−(((1±(√7))/4))=((2−(1±(√7))/4))=((2−1−(√7))/4)=((1−(√7))/4) d=((1−(√7))/4) b=((2−1+(√7))/4)=((1+(√7))/4) =d so a=((1±(√7))/4) b=(((1−(√7))/4) and ((1+(√7))/4)) c=(((1±(√7))/4)) d=(((1−(√7))/4)and ((1+(√7))/4) )...pls check...](https://www.tinkutara.com/question/Q49161.png)
$${ab}−{ad}+{cd}−{bc}=\mathrm{0} \\ $$$${a}\left({b}−{d}\right)−{c}\left({b}−{d}\right)=\mathrm{0} \\ $$$$\left({a}−{c}\right)\left({b}−{d}\right)=\mathrm{0} \\ $$$${either}\:{a}={c}\:\:{or}\:{b}={d} \\ $$$${let}\:{a}={c} \\ $$$${a}^{\mathrm{2}} +{d}^{\mathrm{2}} ={c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:\:\:{so}\:{b}={d} \\ $$$${henc}\:{a}={c}\:\:\:\:{b}={d} \\ $$$${a}+{b}+{c}+{d}=\mathrm{1} \\ $$$$\mathrm{2}{a}+\mathrm{2}{b}=\mathrm{1}\:\:{a}+{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${ac}+{bd}=\mathrm{1} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−{a}\right)^{\mathrm{2}} =\mathrm{1}\:\:\:\left[{a}+{b}=\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}−{a}+{a}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{1}=\mathrm{4} \\ $$$$\mathrm{8}{a}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{3}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{96}}}{\mathrm{16}}=\frac{\mathrm{4}\pm\sqrt{\mathrm{112}}}{\mathrm{16}}=\frac{\mathrm{4}\pm\mathrm{4}\sqrt{\mathrm{7}}}{\mathrm{16}} \\ $$$${a}=\frac{\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{4}}\:\:\:{c}=\frac{\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{2}}−\left({a}\right) \\ $$$$\left.{b}=\frac{\mathrm{1}}{\mathrm{2}}−\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{4}}\right)=\frac{\mathrm{2}−\left(\mathrm{1}\pm\sqrt{\mathrm{7}}\right.}{\mathrm{4}}\right)=\frac{\mathrm{2}−\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}=\frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${d}=\frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${b}=\frac{\mathrm{2}−\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}=\frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}\:={d} \\ $$$${so}\:{a}=\frac{\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${b}=\left(\frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}\:{and}\:\frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}\right) \\ $$$${c}=\left(\frac{\mathrm{1}\pm\sqrt{\mathrm{7}}}{\mathrm{4}}\right) \\ $$$${d}=\left(\frac{\mathrm{1}−\sqrt{\mathrm{7}}}{\mathrm{4}}{and}\:\frac{\mathrm{1}+\sqrt{\mathrm{7}}}{\mathrm{4}}\:\right)…{pls}\:{check}… \\ $$
Commented by behi83417@gmail.com last updated on 03/Dec/18
$${thank}\:{you}\:{so}\:{much}\:{sir}\:{tanmay}. \\ $$$${nice}\:{question}\:{and}\:{beautiful}\:{solution}! \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Dec/18
$${thank}\:{you}\:{sir}.. \\ $$