sloppycanada
  • sloppycanada
How would one go about solving the following rational functions?
Mathematics
katieb
  • katieb
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sloppycanada
  • sloppycanada
|dw:1435629520612:dw|
anonymous
  • anonymous
Equals what again ?
sloppycanada
  • sloppycanada
-2

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sloppycanada
  • sloppycanada
SolomonZelman
  • SolomonZelman
Can you draw your function a little better, please? I can see the left hand side of the equation, \(\large\color{black}{ \displaystyle \frac{5}{5-p}-\frac{p^2}{5-p} }\) but what is this equal to? What is the right side of your equation?
sloppycanada
  • sloppycanada
|dw:1435631324555:dw|
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{5}{5-p}-\frac{p^2}{5-p}=-2}\) combine the two fractions first
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{5-p^2}{5-p}=-2}\) would this be correct, how do you think?
sloppycanada
  • sloppycanada
No this wouldn't be correct? Since -p - (-p) would cancel out right?
SolomonZelman
  • SolomonZelman
it is correct
SolomonZelman
  • SolomonZelman
just as \(\large\color{black}{ \displaystyle \frac{A}{B}-\frac{C}{B}\Rightarrow \frac{A-C}{B}}\) so is \(\large\color{black}{ \displaystyle \frac{5}{5-p}-\frac{p^2}{5-p}\Rightarrow \frac{5-p^2}{5-p}}\)
SolomonZelman
  • SolomonZelman
is this making sense?
sloppycanada
  • sloppycanada
Oh okay.. so then I multiply both sides by the denominator?
SolomonZelman
  • SolomonZelman
yes, you need to multiply times 5-p on both sides
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \frac{5-p^2}{5-p}=-2}\) \(\large\color{black}{ \displaystyle \left(\frac{5-p^2}{5-p}\right)\times (5-p)=-2 \times (5-p)}\)
SolomonZelman
  • SolomonZelman
5-p on the left side cancels \(\large\color{black}{ \displaystyle 5-p^2=-2(5-p)}\)
SolomonZelman
  • SolomonZelman
now, solve for p
sloppycanada
  • sloppycanada
5 - p^2 = -10 + 2p?
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
now, bring everything over to the left side
SolomonZelman
  • SolomonZelman
or, you can subtract 2p from both sides, subtract 5 from both sides, divide by -1 on both sides, and complete the square
sloppycanada
  • sloppycanada
5-p^2 + 10 - 2p = 0 15p-p^2 -2p
SolomonZelman
  • SolomonZelman
15p?
sloppycanada
  • sloppycanada
Sorry 15 - p^2
SolomonZelman
  • SolomonZelman
where is -2p though?
SolomonZelman
  • SolomonZelman
should be: -p²-2p+15=0
sloppycanada
  • sloppycanada
15 - p^2 - 2p = 0?
sloppycanada
  • sloppycanada
oh okay
SolomonZelman
  • SolomonZelman
yes, in my order is going to be more convenient
SolomonZelman
  • SolomonZelman
-p²-2p+15=0 divide both sides by -1, and factor the left side
sloppycanada
  • sloppycanada
wait -5 or 3
SolomonZelman
  • SolomonZelman
you mean the numbers to factor, or solutions for p?
SolomonZelman
  • SolomonZelman
p²+2p-15=0 (p-3)(p+5)=0 so that we have +2p, and yet -15.
SolomonZelman
  • SolomonZelman
now p=?
sloppycanada
  • sloppycanada
p = 3 and -5
sloppycanada
  • sloppycanada
or, sorry I meant or.
SolomonZelman
  • SolomonZelman
yes p=3, p=-5.
SolomonZelman
  • SolomonZelman
this is the original prob, correct?
sloppycanada
  • sloppycanada
Yes.
SolomonZelman
  • SolomonZelman
yw

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