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Find all values $a$ for which there exists an ordered pair (a,b) satisfying the following system of equations:  a + ab^2 & = 40b,  a - ab^2 & = -32b.  List only the values for a.

 May 9, 2019
 #1
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a + ab^2  = 40b

a - ab^2  = -32b       add these

 

2a  = 8b        divide by 2

a  = 4b    (1)

 

Sub (1) into either equation

 

(4b) + (4b)b^2  = 40b

 

4b^3 + 4b - 40b   = 0   

 

4b^3 - 36b = 0

 

4b ( b^2 - 9)  = 0

 

4b ( b + 3) (b - 3)  = 0      set each factor to 0  and solve for b and we have that

 

b =  0        b  = - 3        and b  =  3

 

And a  = 4(0)  = 0         4(-3)  = - 12        4(3)  = 12

 

So....the solutions are   (a, b)  = (0, 0)    (-3, -12)    and  ( 3, 12)

 

cool cool cool

 May 10, 2019

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