Distance formula (converting fraction to quadratic)

[puma072806]

Hello,
The following problem has me a little stumped. I do fine in these sort of problems but for some reason I am stumped on how something was done.

The answer from the book reads as follows:

50/x + 80/x-10 = 2

(Now it gets put into the quadratic formula, which I am not sure how they did. This is where I would like to know how they did it.)

Their answer is as follows.

x²-75x+250=0

How did they get this from the fractions?

 

[JeffM]

Hello,
The following problem has me a little stumped. I do fine in these sort of problems but for some reason I am stumped on how something was done.

The answer from the book reads as follows:

50/x + 80/x-10 = 2

(Now it gets put into the quadratic formula, which I am not sure how they did. This is where I would like to know how they did it.)

Their answer is as follows.

x²-75x+250=0

How did they get this from the fractions?

Please be careful about grouping symbols. You mean 80/(x - 10) not 80/x - 10, which means\(\displaystyle \dfrac{80}{x} - 10,\ not\ \dfrac{80}{x - 10}.\)

In any case, the denis rule is to clear fractions as soon as possible. It's a good rule.

\(\displaystyle \dfrac{50}{x} + \dfrac{80}{x - 10} = 2 \implies\)

\(\displaystyle x(x - 10)\left(\dfrac{50}{x} + \dfrac{80}{x - 10}\right) = x(x - 10)(2) = 2x^2 - 20x \implies\)

\(\displaystyle 50(x - 10) + 80x = 2x^2 - 20x \implies\)

You take it from here.

 

[mmm4444bot]

With these types of equations (sums or differences of algebraic ratios set equal to a number or more of the same), the initial steps are often:

Multiply both sides by the common denominator (shown by JeffM above)

Simplify/rearrange into standard form ax^2+bx+c=0

Divide both sides by greatest common factor of a,b,c (if any), to reduce those coefficients

At this point, you're in good shape to solve, using method(s) that you've learned (factoring, complete the square, taking roots, quadratic formula, et cetera).