##### Determining all real numbers in a rational equation

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Determine all real numbers p and q such that the rational equation (x-p) over x-1 = x-2 over x-q has a solution all real #s except x=1 and x=q, we want to find p and q that make this equation a tautology.

Given two quadratic polynomials ax^2+bx+c = ax^2+bx+y, only if all coefficients of like terms are equal. i.e a=a b=B and c=y

I have cross multiplied them so far and have x squared -9x -px +9p = x squared -2x -x +2. I need help on where to go from here please

Sep 28th, 2015

You've got the right first step, which is to cross multiply!

The 9 that I'm seeing in your first step is actually a q (lowercase Q) in the original problem, which means it's a variable.

So we have x^2 - qx - px + pq = x^2 -3x + 2 .

If you subtract x^2 from both sides, then the x^2's go away altogether:

-qx - px + pq = -3x + 2

You want the two sides to sort of look like each other so that you can figure out what p and q to be, so I would try to get the left side to say ___x + ____ so that it matches the right side: -3x + 2 .

This means you need to factor x out of the first half of the left side of the equation:

x(-q - p) + pq = -3x + 2

Based on the second sentence in the directions of the problem, you know that the coefficients of x and the constants need to equal each other:

-q - p = -3   and  pq = 2

At this point, you would need to choose an equation and solve for a variable and then substitute what you get into the other equation. I would choose the first equation and solve for p:

-q - p = -3

-p = -3 + q

p = 3 - q

Substitute for p into the second equation:

pq = 2

(3-q)q = 2

3q - q^2 = 2

-q^2 - 3q - 2 = 0 or

q^2 + 3q + 2 = 0

You could use the quadratic formula here or factor. I'll factor.

(q+1)(q+2) = 0

q = -1 and q = -2

Now we know what numbers q can be, so you should plug each one separately into one of your equations (like p = 3 - q)  Remember that each specific q value goes with a specific p value.

p = 3 - q

p = 3 - (-1)

p = 4

So p = 4 and q = -1 is a solution to the original equation.

Also, we have

p = 3 - q

p = 3 - (-2)

p = 5

So p = 5 and q = -2 is also a solution to the original equation.

Those are the values for p and q that would make the original rational equation a tautology.

I hope this helps you move forward on the problem! Let me know if you need any more clarification!
Sep 29th, 2015

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Sep 28th, 2015
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Sep 28th, 2015
Oct 18th, 2017
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