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

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