+0  
 
0
999
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avatar+2499 


                                                                                      

 Mar 10, 2016

Best Answer 

 #3
avatar
+5

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

 Mar 10, 2016
 #1
avatar+333 
+5

just solve the quadratic equation below, then solve for  a and b

 Mar 10, 2016
 #2
avatar+5265 
+5

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.

rarinstraw1195  Mar 10, 2016
 #3
avatar
+5
Best Answer

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

Guest Mar 10, 2016
 #4
avatar+33661 
+5

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

 

.

 Mar 10, 2016
 #5
avatar+2499 
0

i understood thank you guys !

 Mar 10, 2016
 #6
avatar+129899 
0

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

 

 

cool cool cool

 Mar 10, 2016

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