Solve for A:
5/(x^2+6 x+8) = A/(x+2)+B/(x+4)
5/(x^2+6 x+8) = A/(x+2)+B/(x+4) is equivalent to A/(x+2)+B/(x+4) = 5/(x^2+6 x+8):
A/(x+2)+B/(x+4) = 5/(x^2+6 x+8)
Subtract B/(x+4) from both sides:
A/(x+2) = 5/(x^2+6 x+8)-B/(x+4)
Multiply both sides by x+2:
Answer: |A = (5 (x+2))/(x^2+6 x+8)-(B (x+2))/(x+4)
Solve for B:
5/(x^2+6 x+8) = A/(x+2)+B/(x+4)
5/(x^2+6 x+8) = A/(x+2)+B/(x+4) is equivalent to A/(x+2)+B/(x+4) = 5/(x^2+6 x+8):
A/(x+2)+B/(x+4) = 5/(x^2+6 x+8)
Subtract A/(x+2) from both sides:
B/(x+4) = 5/(x^2+6 x+8)-A/(x+2)
Multiply both sides by x+4:
Answer: |B = (5 (x+4))/(x^2+6 x+8)-(A (x+4))/(x+2)
IF YOU WISH TO SOLVE FOR NUMERICAL VALUES OF A AND B, THEN SOLVE FOR X FIRST AND SUBSTITUTE:
Solve for x:
x^2+6 x+8 = 0
The left hand side factors into a product with two terms:
(x+2) (x+4) = 0
Split into two equations:
x+2 = 0 or x+4 = 0
Subtract 2 from both sides:
x = -2 or x+4 = 0
Subtract 4 from both sides:
Answer: |x = -2 or x = -4
+5265 A+B=5 and the quadratic equation in the bottow is what you get if you find the equation of the two denominators.
PROOF FOR THAT STATEMENT:
(x+2)(x+4)
Use the FOIL method.
x2+2x+4x+8
Combine Like Terms
x2+6x+8
The only problem is that that'd mean that A*B must equal 5 and A+B must equal 5. Someone else needs to take it from here, either that, or I'm overcomplicating things.
Solve for A:
5/(x^2+6 x+8) = A/(x+2)+B/(x+4)
5/(x^2+6 x+8) = A/(x+2)+B/(x+4) is equivalent to A/(x+2)+B/(x+4) = 5/(x^2+6 x+8):
A/(x+2)+B/(x+4) = 5/(x^2+6 x+8)
Subtract B/(x+4) from both sides:
A/(x+2) = 5/(x^2+6 x+8)-B/(x+4)
Multiply both sides by x+2:
Answer: |A = (5 (x+2))/(x^2+6 x+8)-(B (x+2))/(x+4)
Solve for B:
5/(x^2+6 x+8) = A/(x+2)+B/(x+4)
5/(x^2+6 x+8) = A/(x+2)+B/(x+4) is equivalent to A/(x+2)+B/(x+4) = 5/(x^2+6 x+8):
A/(x+2)+B/(x+4) = 5/(x^2+6 x+8)
Subtract A/(x+2) from both sides:
B/(x+4) = 5/(x^2+6 x+8)-A/(x+2)
Multiply both sides by x+4:
Answer: |B = (5 (x+4))/(x^2+6 x+8)-(A (x+4))/(x+2)
IF YOU WISH TO SOLVE FOR NUMERICAL VALUES OF A AND B, THEN SOLVE FOR X FIRST AND SUBSTITUTE:
Solve for x:
x^2+6 x+8 = 0
The left hand side factors into a product with two terms:
(x+2) (x+4) = 0
Split into two equations:
x+2 = 0 or x+4 = 0
Subtract 2 from both sides:
x = -2 or x+4 = 0
Subtract 4 from both sides:
Answer: |x = -2 or x = -4
+33661 We can write \(\frac{A}{x+2}+\frac{B}{x+4} \rightarrow \frac{(A+B)x+4A+2B}{x^2+6x+8}\)
For this to equal \(\frac{5}{x^2+6x+8}\) we must have A+B = 0 (and 4A+2B = 5)
So A+B = 0
.
+129852 5 / [x^2 + 6x + 8] = A / [x + 2 ] + B / [ x + 4]
On the right, we can get a common denominator and write
[ A ( x + 2)] + [ B(x + 4)] = 5
(A + B)x + 2A + 4B = 5
Equating co-efficients, we have
A + B = 0 → B = -A (1)
2A + 4B = 5 (2)
And subbing (1) into (2), we have
2A + 4(-A) = 5
-2A = 5 → A = -5/2 = -2.5
And, using (1), B = 2.5
So......A + B = 0
