How do you solve the quadratic $4 - \frac{1}{x} = \frac{3}{x^{2}}$ $4-1/x=3/x^2$ using any method?

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How do you solve the quadratic $4 - \frac{1}{x} = \frac{3}{x^{2}}$ $4-1/x=3/x^2$ using any method?

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Answer 1 · 2016-12-27T16:52:07

Get rid of the denominators by multiplying the whole equation by $x^{2}$ $color(blue)(x^2)$

$4 \times x^{2} - \frac{1}{x} \times x^{2} = \frac{3}{x^{2}} \times x^{2}$ $4 color(blue)( xx x^2) - 1/x color(blue)(xx x^2) = 3/x^2 color(blue)(xx x^2)$

$4 x^{2} - x = 3 \leftarrow$ $4x^2 -x = 3" "larr$ make equal to 0

$4 x^{2} - x - 3 = 0$ $4x^2 -x-3 =0$

Find factors of 4 and 3 which subtract to give 1.

$(4 x + 3) (x - 1) = 0 \leftarrow$ $(4x+3)(x-1) =0" "larr$ either factor could be =0

If $4 x + 3 = 0 \to 4 x = - 3 \to x = - \frac{3}{4}$ $4x+3 = 0" "rarr 4x=-3" "rarr x = -3/4$

If $x - 1 = 0 x = 1$ $x-1=0" " x = 1$

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How do you solve the quadratic $4 - \frac{1}{x} = \frac{3}{x^{2}}$ $4-1/x=3/x^2$ using any method?

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How do you solve the quadratic $4 - \frac{1}{x} = \frac{3}{x^{2}}$ $4-1/x=3/x^2$ using any method?