Respuesta :

LammettHash

It's a quadratic equation in disguise. If you let [tex]y=x^{-1}[/tex], then [tex]y^2=x^{-2}[/tex], and we can rewrite the equation as

[tex]5y^2-17y+6=0[/tex]

Solve for [tex]y[/tex] however you like; we get [tex]y=\dfrac25[/tex] and [tex]y=3[/tex].

But we want to solve for [tex]x[/tex], so we have

[tex]y=x^{-1}=\dfrac25\implies x=\dfrac52[/tex]

[tex]y=x^{-1}=3\implies x=\dfrac13[/tex]

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