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show your work for 2-(4/(x^2)) = -(2/x)
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\[2-\frac{4}{x^{2}} = -\frac{2}{x}\] \[\frac{2x^{2}-4}{x^{2}}=\frac{-2x}{x^{2}}\]... x NOT equal to 0 so.... \[2x^{2}-4=-2\] \[x^{2}+x-2=0\]... solve the quadratic equation for x... we get x=-2...x=1
why did you put x^2 after 2 in the beginning of the equation and x after -2 at the end of the equation?
and why did you make the x under -2 into x^2?
sry the third equation should be... \[2x^{2}-4=-2x\.... as for you question.....multiply and divide by x......so as to make the denominator equal on both sides so we can cancel afterwards....
thank you!
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