HELP ASAP!!!! PLEASE! Solving Rational Equations!!!

Question

We're solving rational equations and we get extra credit if we have this problem answered by tomorrow!

 

x+5/x^2-6x-27+x-17/x^2-18x+77=x+1/x^2-14x+33

 

Please include all of your work! 

 

My teacher said there is 3 answers. I've tried working this problem out and can't seem to make the numbers work. 

 

Please help asap!!!!!!!!!!!!

Answer 1

Yes, there are three answers. Unfortunately, the three answers are irrational and so they must be solved numerically. I used an excel spreadsheet to do the number crunching and an online numerical method which I will mention at the appropriate time. Moreover, your teacher stated no restrictions on the use of technology. For someone with a lot of patience and time to spare, this was in fact an excellent extra credit problem.

 

As suggested by Kenneth, all of the denominators can be factored.

Doing so produces the following equation:

 

(x+5)/{(x-9)(x+3)} - 17/{(x-11)(x-7)} = (x+1)/{(x-11)(x-3)} where, as suggested, the product of binomials in curley braces { } appear in the denominator of each fraction when factored.

 

Next, as suggested by Kenneth, the LCD is: (x-9)(x+3)(x-3)(x-11)(x-7)

 

Multiplying the entire equation by this whopping LCD of 5 binomials will cause the denominators to cancel out provided that none of them are zero, which as Kenneth suggested, we must verify at the end.

 

The equation becomes:

 (x+5)(x-3)(x-11)(x-7) - 17(x-9)(x+3)(x-3) = (x+1)(x-9)(x+3)(x-7)

 

Now comes the challenging task of multiplying all of these binomials together to get a polynomial on each side of the equation. The (x-3) on the left side can be factored out, making this tedious process a bit easier. Leaving the right side alone for now and concentrating on the left side, factoring out (x-3) results in:

    (x-3) {(x+5)(x-11)(x-7) - 17(x-9)(x+3)} =

   (x-3) {(x^2-6x-55)(x-7) - 17(x^2-6x-27)} =

  (x-3) { x^3 - 6x^2-55x-7x^2+42x+385 - 17x^2 + 102x + 459 } =

 (x-3) { x^3 - 30x^2 + 89 x + 844 } =

 x^4 - 30x^3 + 89x^2 + 844 x - 3x^3 + 90x^2 -267x - 2532 =

x^4 -33x^3+179x^2 + 577x -2532

 

 

Now on right hand side: (x+1)(x-9)(x+3)(x-7) = (x^2-8x-9)(x^2-4x-21)

                                                                   = x^4 - 4x^3 - 21x^2 - 8x^3 + 32x^2 + 168x -9x^2 + 36x + 189

                                                                   = x^4 - 12x^3 + 2x^2 + 204x + 189

 

The good news is that the x^4 will cancel out from both sides. This results in a third order equation which does in fact have three roots. Notice on the left side the cubic term has coefficient -33 while on the right side the cubic term is -12. So in order to keep the leading term positive, every term on the left will move to the right, changing signs in the process. The resulting cubic equation is:

 

 21x^3 - 177x^2 - 373 x + 2721 = 0

 

 

 The annoying thing here is that all of these coefficients divide by 3 except 373. Otherwise we could divide it out.

 

Now, appealing to the rational root theorem, P, the set of prime factors of 2721 are p={1,3,907}

 while Q, the set of prime factors of 21 are simply {1,3,7} 

So the set of possible candidates for rational roots are P/Q = {1,1/3,1/7,3,3/7,907,907/3,907/7} U

                                                                                        {-1,-1/3,-1/7,-3,-3/7,-907,-907/3,-907/7}

 

Technology is used to verify that 907 is prime. MS Excel was used to check to see if any of these 16 possible

candidates will make this cubic polynomial equal to zero, and alas, NONE OF THEM WORK!!!!

 

This means that the equation does not have any RATIONAL roots. So now the task becomes more complicated 

of finding two different numbers for X where this polynomial changes sign from positive to negative or vice versa.

At this point, examining the graph of this cubic function may prove more efficient for finding where the irrational roots are. Many graphing utilities have a "zoom in" feature, so as to get a good approximation.

 

It turns out that this function changes sign between the values of x=3 and x=4. I used a numerical method called the bisection method to get an approximation of the first solution, correct to 18 digits, as x=3.677461822271063432.

Per the excel spreadsheet, when this value is plugged into the original equation, both sides come out to -0.9429.

 

Now to find the other two solutions, synthetic division must be done numerically to find the coefficient of the

quadratic equation when this first solution is factored out. This was also done using the excel spreadsheet.

 

The quadratic equation that results from factoring out (x-3.67746182...) from  21x^3 - 177x^2 - 373 x + 2721

using synthetic division per the spreadsheet is:

    21x^2 - 99.7733x - 739.913=0, which is correct to 3 digits

 

Finally per the quadratic formula with A=21, B=99.7733, and C = -739.913

 

 the other two roots are: x=8.769082 and x=-4.01797

 

when these are plugged into the original equation they produce -0.75904 and -0.02863 

on both sides, respectively.

 

As suggested, it wouldn't hurt to verify these values with a function plotter or graphic calculator.

I have saved the spreadsheet, if you are interested, but it is not very organized and you will

have a very hard time following it. The problem is already sufficiently complicated.