Math problem

Quin

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I took a math test today, here was the bonus question:

Solve for Q & R (a real number)

(x+4)/(x+1)(x+2)=(Q/x+1)+(R/x+2)

Anyone know how to solve this? Actual numerical values btw, with no variables in them. Good Luck
 

suprarx7nut

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You need to get it into a system....

**EDIT
From the start....

Multiply the entire equation by (x+1)(x+2) resulting in

x+4 = Q(x+2) + R(x+1) and then distributing...

x+4 = Qx + Q2 + Rx +R

**END OF EDIT

Qx + Rx = x => Q + R = 1

2Q + R = 4

now solve using a matrix or just substitution... Or just look at it until you figure it out ;)

Q = 3, R = -1
 
Last edited:

Quin

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How did you break apart 2Q+Qx+Rx+R=x+4? We haven't learned that yet. Btw, your solution doesn't work. R would have to be -2 if Q=3. And wouldn't there still be an infinite number of solutions? Assuming breaking down 2Q+Qx+Rx+R=x+4 in the way you did it is possible, there isn't a limit to the number of solutions. In the equation you made, 2Q+R=4, Q could be 50, R would just have to be -96
 

Big Wang Bandit

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p575169_1.jpg
 

annoyingrob

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Quin said:
How did you break apart 2Q+Qx+Rx+R=x+4? We haven't learned that yet. Btw, your solution doesn't work. R would have to be -2 if Q=3. And wouldn't there still be an infinite number of solutions? Assuming breaking down 2Q+Qx+Rx+R=x+4 in the way you did it is possible, there isn't a limit to the number of solutions. In the equation you made, 2Q+R=4, Q could be 50, R would just have to be -96
His answer is wrong, but the process is right. you can take 2Q+Qx+Rx+R=x+4, and turn it into 2 equations.

factor together all of the X terms, and all of the non-x terms.

Qx + Rx = x

2Q + R = 4

What you end up with is Q+R=1, and 2Q+R=4

Then he put them into a matrix, and reduced it.
1 1 1
2 1 4

1 1 1
1 0 3

0 1 -2
1 0 3

So Q is 3, R is -2

Or if you don't understand matricies, you can use a system of equations.
 

SupraDerk

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suprarx7nut said:
You need to get it into a system....

**EDIT
From the start....

Multiply the entire equation by (x+1)(x+2) resulting in

x+4 = Q(x+2) + R(x+1) and then distributing...

x+4 = Qx + Q2 + Rx +R

**END OF EDIT

Qx + Rx = x => Q + R = 1

2Q + R = 4

now solve using a matrix or just substitution... Or just look at it until you figure it out ;)

Q = 3, R = -1

I think your R is wrong.

You can use the Heaviside cover up method to solve this. It looks like a partial fraction expansion... so

(x+4)/(x+1)(x+2)=(Q/x+1)+(R/x+2)

evaluate the left hand side where the x+2 = 0 (x=-2) to get your R

(-2+4)/(-2+1) = -2 = R


now find Q

evaluate the left hand side where the x+1 = 0 (x=-1) to get your Q

(-1+4)/(-1+2) = 3 = Q


---------------------

(x+4)/(x+1)(x+2)=(3/x+1)+(-2/x+2)

X = 1 ... 5/(2*3)= 5/6
(3/2) - (2/3) = 5/6
 

suprarx7nut

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SupraDerk said:
I think your R is wrong.

You can use the Heaviside cover up method to solve this. It looks like a partial fraction expansion... so

(x+4)/(x+1)(x+2)=(Q/x+1)+(R/x+2)

evaluate the left hand side where the x+2 = 0 (x=-2) to get your R

(-2+4)/(-2+1) = -2 = R


now find Q

evaluate the left hand side where the x+1 = 0 (x=-1) to get your Q

(-1+4)/(-1+2) = 3 = Q


---------------------

(x+4)/(x+1)(x+2)=(3/x+1)+(-2/x+2)

X = 1 ... 5/(2*3)= 5/6
(3/2) - (2/3) = 5/6


Ya, my bad. I didn't go back to check it. And I was too lazy to make the matrices in my calc, so I just eye-balled it. R is obviously -2, I was hallucinating.
 

suprarx7nut

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Haha, ok so that could easily be done by hand like you did, but I bet I could enter it in my calc before you could do it by hand.

I spent a semester doing matrices in Diff Eq so I'm pretty quick now. I bet I could do it with my eyes closed.
 

Quin

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Dec 5, 2006
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SupraDerk said:
I think your R is wrong.

You can use the Heaviside cover up method to solve this. It looks like a partial fraction expansion... so

(x+4)/(x+1)(x+2)=(Q/x+1)+(R/x+2)

evaluate the left hand side where the x+2 = 0 (x=-2) to get your R

(-2+4)/(-2+1) = -2 = R


now find Q

evaluate the left hand side where the x+1 = 0 (x=-1) to get your Q

(-1+4)/(-1+2) = 3 = Q


---------------------

(x+4)/(x+1)(x+2)=(3/x+1)+(-2/x+2)

X = 1 ... 5/(2*3)= 5/6
(3/2) - (2/3) = 5/6

Can you explain that better? We haven't covered that in my class... :(
 

SupraDerk

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This place does a pretty decent job of explaining and showing:
http://cnx.org/content/m2111/latest/

Are you in Calc 2? If so, then you'll probably get penalized for using this method as they usually want to make you go the long ass route so you can see what's going on.

I learned this method in my Signals and Linear Systems class, at that point the professor was only interested in helping us get to our answer as fast as possible
 

Quin

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Thanks, that's pretty damn sweet actually. I'm in high school (H) Algebra II, so I'm pretty sure this question was way out of our league, but I learned how to do it so its all good. Thanks guys