Question 18

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- Jul 12th 2018, 08:15 AM #1

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- Jul 12th 2018, 01:46 PM #2

- Jul 12th 2018, 04:03 PM #3

- Jul 12th 2018, 06:07 PM #4

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- Jul 12th 2018, 06:51 PM #5
## Re: Solve for 1/x

Did you study this HERE ?

If not then you are lazy.

If you did you know that $\large x=\dfrac{b}{a}-1=\dfrac{b-a}{a},~a\ne 0$ so $\large\dfrac{4}{x}=\dfrac{4a}{b-a}$

You can show that $\large 1+\dfrac{4}{x}+\dfrac{4}{x^2}=?$

the answer is (2) HOW???

- Jul 13th 2018, 04:49 AM #6

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- Jul 13th 2018, 05:56 AM #7

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## Re: Solve for 1/x

$$abx^2=(a-b)^2(x+1)$$

This is a quadratic in $x$:

$$(ab)x^2 - (a-b)^2 x - (a-b)^2 = 0$$

$$\begin{align*}x & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^4 + 4ab(a-b)^2}}{2ab} \\ & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^2[(a-b)^2+4ab]}}{2ab} \\ & = \dfrac{(a-b)^2 \pm \sqrt{(a-b)^2(a+b)^2}}{2ab} \\ & = \dfrac{(a-b)^2 \pm |a-b||a+b|}{2ab} \\ & = \dfrac{(a-b)^2 \pm |a^2-b^2|}{2ab}\\ & = \dfrac{(a-b)^2 \pm (a^2-b^2)}{2ab}\end{align*}$$

Can you finish from there?

- Jul 13th 2018, 06:20 AM #8

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- Jul 13th 2018, 08:26 AM #9

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