SOLUTION: Real numbers a and b satisfy a + ab^2 = 250b a + b = 102 Enter all possible values of a, separated by commas.

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Question 1209082: Real numbers a and b satisfy
a + ab^2 = 250b
a + b = 102
Enter all possible values of a, separated by commas.
Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!

system%28a+%2B+ab%5E2+=+250b%2C%0D%0Aa+%2B+b+=+102%29

Solve the second for b = (102-a) and substitute in the first:

a+%2B+a%28102-a%29%5E2+=+250%28102-a%29

a+%2B+a%2810404-204a%2Ba%5E2%29=25500-250a

a+%2B+10404a-204a%5E2%2Ba%5E3=25500-250a

a%5E3-204a%5E2%2B10655a-25500=0

Since the last term is divisible by 100, let's take a chance and
see if 100 is a root with synthetic division:

100 | 1 -204  10655 -25500
    |    100 -10400  25500
      1 -104    255      0

Sure enough, 100 is a root! So the cubic factors as

%28a-100%29%28a%5E2-104a%2B255%29=0
a-100=0;  a2-104a+255=0
    a=100; 
  b=102-a;
b=102-100=2

So there is one solution: a=100, b=2. (a,b) = (100,2)

Checking:
system%28a+%2B+ab%5E2+=+250b%2C%0D%0Aa+%2B+b+=+102%29
system%28100+%2B+%28100%29%282%29%5E2+=+250%282%29%2C%0D%0A100+%2B+2+=+102%29
system%28100+%2B+400+=+500%2C%0D%0A102=+102%29
That checks.  But that cubic equation has too more solutions.

You know how to solve a quadratic by the quadratic formula,

So solve this quadratic equation by the quadratic formula as
it doesn't factor:

a%5E2-104a%2B255=0

and you'll get two more solutions for " a ",

Then substitute them in b = 102-a and you'll have two more 
solutions (a,b).

Edwin