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Find all values \(a\) for which there exists an ordered pair \((a,b)\) satisfying the following system of equations:

\(\begin{align*} a + ab^2 & = 40b, \\ a - ab^2 & = -32b. \end{align*}\)

 Jun 29, 2019
 #1
avatar+8720 
+5

Add the two equations together to get

 

2a  =  8b

a  =  4b

 

Substitute this value in for  a  in the first given equation.

 

4b + (4b)b2  =  40b

 

4b + 4b3  =  40b

                            Subtract  40b  from both sides of the equation.

4b3 - 36b  =  0

                            Factor  b  out of the terms on the left side.

b(4b2 - 36)  =  0

                                     Factor  4b2 - 36  as a difference of squares.

b(2b + 6)(2b - 6)  =  0

                                     Set each factor equal to zero and solve for  b

 

b  =  0 ____or____ 2b + 6  =  0 ____or____ 2b - 6  =  0
    2b  =  -6

 

 

2b  =  6
    b  =  -3   b  =  3

 

If     b  =  0     then     a  =  4b  =  4(0)  =  0          so     (0, 0)  is a solution.

 

If     b  =  -3     then     a  =  4b  =  4(-3)  =  -12          so     (-12, -3)  is a solution.

 

If     b  =  3     then     a  =  4b  =  4(3)  =  12          so     (12, 3)  is a solution.

 Jun 29, 2019
 #2
avatar+59 
+1

thank you!!!

 Jun 29, 2019

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