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So I've got this quadratic equation and am totally unable to solve it. Can someone tell me how to do it?

$$\frac{a}{ax-1} + \frac{b}{bx-1} = a + b,$$

where $x$ is not equal to $\frac{1}{a}$ or $\frac{1}{b}$. We need to solve for $x$.

Casteels
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Bone
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    Multiply both sides of your equation by $(ax - 1)(bx - 1)$, expand, collect terms on one the LHS and use the quadratic formula.. – Matthew Cassell Mar 15 '15 at 14:08
  • looks like $x = \frac 1{a+b}, x = \frac{a^2 + ab + b^2}{ab(a+b)}$ – abel Mar 15 '15 at 14:45
  • How did you get that? I'm unable to solve it even after the answer given below. – Bone Mar 15 '15 at 14:46
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    @String I'm not laughing. That's a stupid comment. Provide the link, but don't pretend you're offering tips for proceeding with the problem. Save your efforts for users who've been around awhile (at least more than a day!) – amWhy Mar 15 '15 at 14:49
  • i will post the answer. @amWhy, good work. – abel Mar 15 '15 at 14:53
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    OK, sorry guys! The original post read $$(a/ax-1)+(b/bx-1)=a+b$$ which simplifies to the non-quadratic $2x-2=a+b$. So may I suggest the OP either to write $$a/(ax-1)+b/(bx-1)=a+b$$ or to consult this page for more advanced formatting. Also you can right-click on TeX-formatted formulas written by others and choose Show Math As>TeX Commands to learn more. – String Mar 15 '15 at 15:03

3 Answers3

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$$\frac{a}{ax-1}+\frac{b}{bx-1}{=a+b} \Rightarrow \frac{a(bx-1)+b(ax-1)}{(ax-1)(bx-1)}=a+b \Rightarrow \frac{abx-a+abx-b}{(ax-1)(bx-1)}=a+b \Rightarrow \frac{2abx-a-b}{(ax-1)(bx-1)}=a+b \Rightarrow 2abx-a-b=(ax-1)(bx-1)(a+b)$$

Can you continue?

evinda
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$$\frac{a}{a x -1}-b =a-\frac{b}{b x-1}$$

$$\frac{a b x-a-b}{a x-1} +\frac{a b x-a - b}{b x-1}=0$$

$$(a b x-a-b)\left(\frac{1}{a x-1} +\frac{1}{b x-1}\right)=0$$

$$(a b x-a-b)((a+b)x-2 )=0$$

$$x=\left\{\frac{2}{a+b},\frac{a+b}{a b}\right\}$$

Lozenges
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if you multiply the equation by $(ax-1)(bx-1),$ you get $$abx -a + abx - b = (a+b)\left(abx^2 - (a+b)x + 1\right)$$ you simplify this to $$ab(a+b)x^2 -x\left(a^2+ 4ab + b^2\right) + 2(a+b) = 0$$ the discriminant is

$$\begin{align} disc &= \left(a^2+ 4ab + b^2\right)^2 - 8ab(a+b)^2 \\ &= a^4 + 16a^2b^2 + b^4 + 8a^3b + 2a^2b^2 + 8ab^3 - \left( 8ab(a^2+2ab+b^2) \right)\\&= (a^2 + b^2)^2\end{align}$$

now, $$x = \frac{a^2 + 4ab + b^2 \pm(a^2 + b^2)}{2ab(a+b)}=\frac{a+b}{ab}, \frac 2{a+b} $$

abel
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