The Quadratic Formula: 2/(X^2+6X+8)+3/(X^2+5X+4)=X/(X^2+3X+2

[helphelphelphelp]

Q: 2/(X^2 + 6X +8) + 3/(X^2 + 5X + 4) = X/(X^2 + 3X +2)


What I did: Cross multiplied and got rid of the fractions. Collected the liketerms

The Problem: Got this answer : X^5 + 6X^4 - X^3 - 62X^2 - 120X - 64
(when there is supposed to be Two answers that are both Real Numbers) What this section requires me to do is to basically factor out using the Quadratic Formula. It's a little bit difficult because of the fact that I have 6 terms instead of the usual 3.

Thanks for the help!

 

[Loren]

Re: The Quadratic Formula

First, get each of the denominators in factored form. Then determine the least common denominator and multiply both sides of the equation by that LCD. If you do that, I think your work will be substantially easier.

 

[Denis]

Re: The Quadratic Formula

As Loren said, FACTOR the denominators; you get this:
2 / [(x+4)(x+2)] + 3 / [(x+4)(x+1)] = x / [(x+2)(x+1)]

Now you should find it easy to directly end up with a quadratic.

 

[helphelphelphelp]

great thx alot!

 

[Deleted member 4993]

Q: 2/(X^2 + 6X +8) + 3/(X^2 + 5X + 4) = X/(X^2 + 3X +2)


What I did: Cross multiplied and got rid of the fractions. Collected the liketerms <<< This is why cross-multiplication should not be discussed before graduate school.

The Problem: Got this answer : X^5 + 6X^4 - X^3 - 62X^2 - 120X - 64 <<< Where did the '[sub:3nqfz7ma]=[/sub:3nqfz7ma]' sign go?