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Fool's problem of the day, If \(2x^4+x^3-11x^2+x+2=0\), what are the values of \(\large ( x+\frac1x )\) ?
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we know x=0 is not a valid solution. therefore:\[\begin{align} 2x^4+x^3-11x^2+x+2&=0\\ (2x^2-5x+2)(x^2+3x+1)&=0\\ \end{align}\]therefore:\[2x^2-5x+2=0\tag{a}\]or:\[x^2+3x+1=0\tag{b}\] From (a) we get:\[x(2x-5+\frac2x)=0\]\[x(2(x+\frac1x)-5)=0\]therefore, x=0 (reject as this is not valid), or:\[2(x+\frac1x)-5=0\implies x+\frac1x=\frac{5}{2}\] From (b) we get:\[x(x+3+\frac1x)=0\]\[x(x+\frac1x+3)=0\]therefore, x=0 (reject as this is not valid), or:\[x+\frac1x+3=0\implies x+\frac1x=-3\]
Awesome!
Nice Asnaseer, but I would have just divided the whole thing by \(x^2\)
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