2

What is the common denominator of the following:

$x^2-x$ and $2x$, $x^2$

(the first is the one side, and the other 2 are another side of the formula)

The full formula is: $1/(x^2-x) = 1/2x + 1/x^2$

P.G

That's not homework. I am a mother of 3 trying to enjoy math again

Mankind
  • 13,170
Dejel
  • 139
  • I don't see a common denominator tag and I can't add a new one with my reputation – Dejel Oct 24 '15 at 20:13
  • The common denominator of $\frac{1}{2x}$ and $\frac{1}{x^2}$ is $2x^2$; that is what you need here. – vadim123 Oct 24 '15 at 20:18
  • To clarify: are you asking which $x$ satisfy $\frac {1}{x^2-1}=\frac {1}{2x}+\frac {1}{x^2}$? – lulu Oct 24 '15 at 20:19
  • @vadim123 I need a common one with X^2-x – Dejel Oct 24 '15 at 20:21
  • @lulu, well, I finally would need to reach what x satisfy - but first I need to find the common denominator - no? – Dejel Oct 24 '15 at 20:22
  • 1
    No, you don't. You can do it that way, but I think it is simpler to start by noting that every denominator is divisible by $x$ so multiplying both sides of the equation by $x$ makes life simpler. Then combine the terms on the right and cross multiply. – lulu Oct 24 '15 at 20:25

2 Answers2

3

Your equation is

$$\frac{1}{x^2-x} = \frac{1}{2x} + \frac{1}{x^2}.$$

To find a common denominator, the first thing to look for is if any of the current denominators can be factored. The first one can, as $x^2-x = x(x-1)$, so your equation becomes

$$\frac{1}{x(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$

You then look for what each denominator is missing compared to the others. When I compare the first denominator $x(x-1)$ to the second $2x$, then the first one already has the $x$, but it is missing the factor of $2$, and when I compare the first denominator to the third $x^2$, I see that I am missing one factor of $x$ in the first to match the $x^2$ in the third. In total, I multiply the numerator and denominator of the first fraction by $2x$ to get

$$\frac{2x}{2x^2(x-1)} = \frac{1}{2x} + \frac{1}{x^2}.$$

The second denominator $2x$ needs to change to become equal to the new first denominator, because this is our guess of a common denominator - but to change $2x$ to $2x^2(x-1)$, I need only multiply by $x(x-1)$, because this is what is missing. Similarly, to change the third denominator $x^2$ to $2x^2(x-1)$, I multiply by $2(x-1)$. In total, you equation becomes

$$\frac{2x}{2x^2(x-1)} = \frac{x(x-1)}{2x^2(x-1)} + \frac{2(x-1)}{2x^2(x-1)},$$

and your common denominator is $2x^2(x-1)$.

Mankind
  • 13,170
  • Thanks! how do you format the forumulas so nicely? – Dejel Oct 24 '15 at 20:32
  • No problem. I use something called "TeX commands" that this site supports. You basically use various commands for mathematics and incapsulate it in dollar signs. For instance, \frac{a}{b} is a fraction that gives you $\frac{a}{b}$ when you put dollar signs on each side of it. There's a quick guide here with further links: http://math.stackexchange.com/help/notation Also, you can right click on any of the math in my post and click "Show Math As TeX Commands" to see the underlying commands. – Mankind Oct 24 '15 at 20:37
2

A common denominator is a polynomial that is divisible for all the denominators. Since $x^2-x=x(x-1)$, the simpler such polinomial is $2x^2(x-1)$ that contains all factors in the denominators.

Emilio Novati
  • 62,675