Question-69018

Question Number 69018 by Maclaurin Stickker last updated on 17/Sep/19

Commented by Prithwish sen last updated on 18/Sep/19

It is only possible when 4a+5b+1=8ab+1 ⇒4a+5b=8ab............(i) and (16a^2 +b^2 +1)(4a+5b+1)=(4a+5b+1)^2 =((8ab+1)^2 ......(ii) ⇒ 16a^2 +b^2 +1=4a+5b+1 16a^2 +b^2 +1= 8ab+1 from (i) (4a−b)^2 =0 ⇒4a=b.......(ii) ∴ from (i) and (ii) we get a=0 , b=0 and a=(3/4) and b=3 ∵ the equation only satified by a=(3/4) and b=3 ∴ the possible solution is ((3/4),3)

$$\mathrm{It}\:\mathrm{is}\:\mathrm{only}\:\mathrm{possible}\:\mathrm{when} \\ $$$$\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{5}\boldsymbol{\mathrm{b}}+\mathrm{1}=\mathrm{8}\boldsymbol{\mathrm{ab}}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{5}\boldsymbol{\mathrm{b}}=\mathrm{8}\boldsymbol{\mathrm{ab}}…………\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\boldsymbol{\mathrm{and}} \\ $$$$\left(\mathrm{16}\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{5}\boldsymbol{\mathrm{b}}+\mathrm{1}\right)=\left(\mathrm{4}\boldsymbol{\mathrm{a}}+\mathrm{5}\boldsymbol{\mathrm{b}}+\mathrm{1}\right)^{\mathrm{2}} =\left(\left(\mathrm{8}\boldsymbol{\mathrm{ab}}+\mathrm{1}\right)^{\mathrm{2}} ……\left(\boldsymbol{\mathrm{ii}}\right)\right. \\ $$$$\Rightarrow\:\mathrm{16a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{1}=\mathrm{4a}+\mathrm{5b}+\mathrm{1} \\ $$$$\mathrm{16a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{1}=\:\mathrm{8ab}+\mathrm{1}\:\:\boldsymbol{\mathrm{from}}\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\left(\mathrm{4a}−\mathrm{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}…….\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$\therefore\:\mathrm{from}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$$\boldsymbol{\mathrm{a}}=\mathrm{0}\:,\:\boldsymbol{\mathrm{b}}=\mathrm{0}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{a}}=\frac{\mathrm{3}}{\mathrm{4}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}=\mathrm{3} \\ $$$$\because\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{equation}}\:\boldsymbol{\mathrm{only}}\:\boldsymbol{\mathrm{satified}}\:\boldsymbol{\mathrm{by}}\: \\ $$$$\boldsymbol{\mathrm{a}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}=\mathrm{3}\: \\ $$$$\therefore\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{solution}}\:\boldsymbol{\mathrm{is}}\:\left(\frac{\mathrm{3}}{\mathrm{4}},\mathrm{3}\right) \\ $$

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