0
$\begingroup$

How can I show the following relationship algebraically? $$\frac 1 {4-\frac2 x}=\frac1 4+\frac1 {8x-4}\ \ \ \ \forall x\ne\frac1 2$$ I tried to multiply by the conjuage $$\left(\frac 1 {4-\frac2 x}\right)\left(\frac{4+\frac2 x}{4+\frac2 x}\right)=\frac{2+\frac 1 x}{8-\frac2 {x^2}}$$ and ended up nowhere. I realized that partial fraction decomposition must be a valid solution, but I cannot figure out how I would set it up with only the first fraction as information.

To describe where the problem came from, I divided the polynomial $\frac{x}{4x-2}$ in the wrong order, (that is, I calculated $\frac{4x-2}x$) and I figured I could just take the reciprocal of what I came up with to get the right answer.

$\endgroup$
3
$\begingroup$

$\displaystyle\frac1{4-\frac2x}=\frac{x}{4x-2}=\frac{\frac14(4x-2)+\frac12}{4x-2}=\frac14+\frac{\frac12}{4x-2}=\frac14+\frac1{8x-4}$

$\endgroup$
  • $\begingroup$ Thank you for the quick and insightful answer. Do you think you could express when multiplying a fraction by it's conjugate can be useful? $\endgroup$ – Will Sherwood Aug 4 '16 at 4:55
  • $\begingroup$ When it has surds $\endgroup$ – Kenny Lau Aug 4 '16 at 4:56
  • $\begingroup$ Thank you for your appreciation. $\endgroup$ – Kenny Lau Aug 4 '16 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.