SOLUTION: Given a+b=1, a^3 + b^3=16 and (a+b)^3 = a^3 +3a^2b +3ab^2 + b^3, find the value of a^2 + b^2.

Algebra ->  Equations -> SOLUTION: Given a+b=1, a^3 + b^3=16 and (a+b)^3 = a^3 +3a^2b +3ab^2 + b^3, find the value of a^2 + b^2.       Log On


   

Question 1116399: Given a+b=1, a^3 + b^3=16 and (a+b)^3 = a^3 +3a^2b +3ab^2 + b^3, find the value of a^2 + b^2.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a+b=1
a^3 + b^3=16
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Sub 1-a for b
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a^3 + (1-a)^3 = 16
a^3 + 1 -3a +3a^2 -a^3 = 16
3a^2 - 3a - 15 = 0
a^2 - a - 5 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-1x%2B-5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-1%29%5E2-4%2A1%2A-5=21.

Discriminant d=21 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--1%2B-sqrt%28+21+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+21+%29%29%2F2%5C1+=+2.79128784747792
x%5B2%5D+=+%28-%28-1%29-sqrt%28+21+%29%29%2F2%5C1+=+-1.79128784747792

Quadratic expression 1x%5E2%2B-1x%2B-5 can be factored:
1x%5E2%2B-1x%2B-5+=+%28x-2.79128784747792%29%2A%28x--1.79128784747792%29
Again, the answer is: 2.79128784747792, -1.79128784747792. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-1%2Ax%2B-5+%29

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a = x1, b = x2 or vice versa
You do a^2 + b^2
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You'll find a^2 + b^2 = 11, so there is more than one way to do it.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

The way  (= the ONLY way;  == the CANONICAL way) of solving this problem is  THIS:



1.  1 = %28a%2Bb%29%5E3 = a%5E3+%2B+3a%5E2%2Ab+%2B+3a%2Ab%5E2+%2B+b%5E3 = %28a%5E3%2Bb%5E3%29 + 3ab%28a%2Bb%29 = (replace a^3 + b^3 by 16 and replace a+b by 1, since it is given) = 


        16 + 3ab,

    which implies  3ab = 1 - 16 = -15  and hence  ab = -5.    (*)



2.  Now the next and the last step is  


    1 = %28a%2Bb%29%5E2 = a%5E2+%2B+2ab+%2B+b%5E2 = %28a%5E2%2Bb%5E2%29 + %282ab%29 = (replace ab by -5, since we just found it in (*)) = a%5E2+%2B+b%5E2 - 2%2A5 = a%5E2+%2B+b%5E2 - 10,


    which implies  a%5E2+%2B+b%5E2 = 1 + 10 = 11.


Answer. Under given conditions,  a%5E2+%2B+b%5E2 = 11.