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QRAwarrior
6x + 22 = A(x + 2) + B(x - 4) ^I am doing partial fraction decomposition and this is what I am at. Attempt: "Trick" Let x = -2. Then, 6(2) + 22 = A(-2 + 2) + B(-2 - 4) 12 + 22 = 0 + B(-6) 34 = -6B -34/6 = B ------ "No Trick" 6x + 22 = Ax + 2A + Bx - 4B (LHS = RHS, then coeff's are equal), 6 = A + B, [1] 22 = 2A - 4B => 11 = A - 2B, [2a] [2a] - [1], 5 = 0 -B => -5 = B. Why aren't the two answers the same?
i am lost. where does this line come from \[6x + 22 = Ax + 2A + Bx - 4B\]
oh nvm i see the mistake
6x + 22/(x+2)(x-4) dx
What is the mistake?
Let x = -2. Then, 6(2) + 22 = A(-2 + 2) + B(-2 - 4)
should be Let x = -2. Then, 6(-2) + 22 = A(-2 + 2) + B(-2 - 4)
Ok I actually did it with -2, and then this is what I got: 6x + 22 = A(x + 2) + B(x-4). Let x = -2. Then, 6(-2) + 22 = A(-2 + 2) + B(-2-4) -12 + 22 = B(-6) 10/6 = -B -10/6 = B....
6x+22=(a+b)x+(2a-4b) now comparing both sides 6=a+b 22=(2a-4b) now solve two equations you will get a and b
@SUROJ I did that for "No Trick"
\[a+b=6\] \[a+2b=11\] \[b=5\]
What am I doing wrong wrt the trick? Why am I not getting the same answer as the second method?
(6x+22)/((x+2)(x-4)) Factor out the 2, 2(3x+11)/((x+2)(x-4)) ---------------------------------------- A/(x+2)+B/(x-4) ---------------------------------------- 6=A+B 22=-4A+2B ---------------------------------------- A=-5/3 B=23/3 -(5)/(3(x+2))+(23)/(3(x-4))
i see mistake when you put 6(2) in place of 6(-2) ....in first trick
ok lets start at the top is it \[\frac{6x + 22 }{(x+2)(x-4)}=\frac{a}{x+2}+\frac{b}{x-4}\]
Ok folks, the point here is that the "Trick" does not match with "No Trick" for the below solution: 6x + 22 = A(x + 2) + B(x-4). Let x = -2. Then, 6(-2) + 22 = A(-2 + 2) + B(-2-4) -12 + 22 = B(-6) 10/6 = -B -10/6 = B....
Let x = -2. Then, 6(2) + 22 = A(-2 + 2) + B(-2 - 4) || in this step 12 + 22 = 0 + B(-6) 34 = -6B -34/6 = B
@SUROJ Look at my recent one
Then why is it NOT the same as "No trick"?
ok so we can do this by writing \[6x+22=A(x+2)+B(x-4)\] \[-12+22=-6B\] \[10=-6B\] \[-\frac{5}{3}\] and likewise \[A=\frac{23}{3}\] so give \[\frac{6x+22}{(x+2)(x-4)}=\frac{23}{3(x+2)}-\frac{5}{3(x+4)}\]
a+b=6 is same as 2(a+b)=2(6) is same as 2a+2b=12 is same as -2a -2b =-12 add -2a-2b=-12 and 2a-4b=22
Alright, then how about the "No trick"?
typo on last line should be \[\frac{6x+22}{(x+2)(x-4)}=\frac{23}{3(x+2)}-\frac{5}{3(x-4)}\]
add left left and right right -2a-2b=-12 and 2a-4b=22 gives -6b=10 so b is -10/6 same as first one
now we try with "no trick" \[6x+22=A(x-4)+B(x+2)\] \[6x+22=Ax-4A+Bx+2B\] \[6x+22=(A+B)x+2B-4A\] \[A+B=6\] \[2B-4A=22\]
i get the same answer solving this one