Determining all real numbers in a rational equation
Algebra

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Determine all real numbers p and q such that the rational equation (xp) over x1 = x2 over xq 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
Thank you for the opportunity to help you with your question!
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
(3q)q = 2
3q  q^2 = 2
Solve this quadratic equation:
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!Secure Information
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