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I seem to have trouble with quadratic equations when it comes to fractions and square roots.

$$ \frac{1}{x}+2x=3 $$ How do I solve this equation?

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    $\begingroup$ Hint: $x=0$ can not be a solution, so multiply the whole equation by $x$. $\endgroup$ – dxiv Dec 12 '16 at 18:21
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    $\begingroup$ A general strategy is, as a first step, to clear any fractions by multiplying by the LCD: softschools.com/math/algebra/topics/… $\endgroup$ – Daniel R. Collins Dec 12 '16 at 18:25
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As well mentioned by @dxiv in the comment, you can easily see that $x$ cannot be zero (otherwise in the expression on the left we will do something which is not permitted to do so [which I shall let you find]). So, you can multiply the equation by $x$.

On multiplying whole equation by $x$, you get $1+2x^{2}=3x$ $\implies 2x^2-3x+1=0$. On factorising, it becomes $(2x-1)(x-1)$. So, $x=1/2$ or $x=1$. As required.

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    $\begingroup$ Nice post; but You should say something about why you can multiply by x (that it's not zero). $\endgroup$ – Namaste Dec 12 '16 at 18:27
  • $\begingroup$ You can multiply by x even when it is 0! It just doesn't help solve the equation because then the equation reduces to "0= 0". $\endgroup$ – user247327 Dec 12 '16 at 18:35
  • $\begingroup$ Thanks @amWhy, I have edited it now. $\endgroup$ – I am Back Dec 12 '16 at 18:37
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    $\begingroup$ @user247327 no, because you'd have division by $0$ first, and you can't just multiply that by $0$ expecting anything times zero equals zero. $\endgroup$ – Simply Beautiful Art Dec 12 '16 at 23:59
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Or, multiply by $\, y = x^{-1}$ to get $\ y^2-3y +2 = (y\!-\!2)(y\!-\!1)= 0\ $ so $\, x^{-1} = y = 2,1$

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