1. Solve for 1/x

Question 18

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2. Re: Help #2

Please use a descriptive title for your thread. It helps us (and you for that matter) to keep track of the conversations. "Help #2" tells us nothing.

-Dan

3. Re: Solve for 1/x

Study this partial solution, HERE

4. Re: Solve for 1/x

Originally Posted by sbjsavio
Question 18

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Stuck here

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5. Re: Solve for 1/x

Originally Posted by sbjsavio

Stuck here
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 $\large 1+\dfrac{4}{x}+\dfrac{4}{x^2}=?$

6. Re: Solve for 1/x

Originally Posted by Plato
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 $\large 1+\dfrac{4}{x}+\dfrac{4}{x^2}=?$

I end up getting a-b whole cube divided by ab
And that wolfram alpha is not free,i need to buy premium to view it(not lazy)
I didnt get that how u got x equals

7. 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?

8. Re: Solve for 1/x

rewrite the equation in the form

$\displaystyle \frac{x+1}{x^2}=\frac{a b}{(a-b)^2}$

$\displaystyle \frac{1}{x^2}+\frac{1}{x}=\frac{a b}{(a-b)^2}$

9. Re: Solve for 1/x

Ideas method is easiest.thanks man