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6x^-2 -7x^-1 +1=0

Solve by making appropriate substitutions

Verified Answer :

LammettHash

Let \(y=x^{-1}\), so that \(y^2=x^{-2}\). Then

\(6x^{-2}-7x^{-1}+1=0\iff6y^2-7y+1=0\implies (6y-1)(y-1)=0\)

which has two solutions, \(y=\dfrac16\) and \(y=1\). So the solutions in terms of \(x\) would be

\(x^{-1}=\dfrac16\implies x=6\)
\(x^{-1}=1\implies x=1\)

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