Question Number 20552 by ajfour last updated on 28/Aug/17
$${The}\:{roots}\:{of}\:{the}\:{equation}\: \\ $$$$\:\left(\mathrm{3}−{x}\right)^{\mathrm{4}} +\left(\mathrm{2}−{x}\right)^{\mathrm{4}} =\left(\mathrm{5}−\mathrm{2}{x}\right)^{\mathrm{4}} \:{are} \\ $$$$\left({a}\right)\:{all}\:{real}\:\:\:\:\left({b}\right)\:{all}\:{imaginary} \\ $$$$\left({c}\right)\:{two}\:{real}\:{and}\:{two}\:{imaginary} \\ $$$$\left({d}\right){none}\:{of}\:{the}\:{above}\:. \\ $$
Answered by Tinkutara last updated on 28/Aug/17
$$\left({x}−\mathrm{3}\right)^{\mathrm{4}} +\left({x}−\mathrm{2}\right)^{\mathrm{4}} =\left(\mathrm{2}{x}−\mathrm{5}\right)^{\mathrm{4}} \\ $$$${Let}\:{a}={x}−\mathrm{3},\:{b}={x}−\mathrm{2},\:{then}\:{solving}\:{for} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\left({a}+{b}\right)^{\mathrm{4}} \\ $$$${ab}\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{ab}\right)=\mathrm{0} \\ $$$${One}\:{solution}\:{is}\:{ab}=\mathrm{0}\Rightarrow\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{2},\mathrm{3}\:\left({Two}\:{integral}\:{roots}\right) \\ $$$${and}\:\mathrm{2}\left({a}+{b}\right)^{\mathrm{2}} ={ab} \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{2}{x}−\mathrm{5}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$$$\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{20}{x}+\mathrm{25}\right)={x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6} \\ $$$$\mathrm{7}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{44}=\mathrm{0} \\ $$$${D}=\mathrm{1225}−\mathrm{4}×\mathrm{7}×\mathrm{44}=−\mathrm{7}<\mathrm{0} \\ $$$${Hence}\:\mathrm{2}\:{real}\:{and}\:\mathrm{2}\:{imaginary}\:{roots}. \\ $$
Commented by ajfour last updated on 28/Aug/17
$${Excellent}!\:{keep}\:{it}\:{up}. \\ $$