+0  
 
0
2
1
avatar+432 

Find one ordered pair $(x,y)$ of real numbers such that $x + y = 10$ and $x^3 + y^3 = 162 + x^2 + y^2.$

 Jun 10, 2024
 #1
avatar+88 
+1

You might want to double check the question, because there is no real pair of numbers that holds those properties. There are, however, complex numbers that do:

 

\(y = 10 - x\)

\(x^3 + (10-x)^3 = 162 +x^2 + (10-x)^2\)

\(x^3 + (10-x)^2(10-x) = 162 + x^2 + (100 - 20x + x^2)\)

\(x^3 + (100-20x+x^2)(10-x) = 2x^2-20x+262\)

\(x^3 + (1000 -200x+10x^2-100x+20x^2-x^3) = 2x^2-20x+262\)

\(30x^2-300x+1000 = 2x^2 - 20x + 262\)

\(15x^2-150x+500=x^2-10x+131\)

\(14x^2-140x+369 = 0\)

\(x^2-10x+\frac{369}{14} = 0\)

\(x^2 - 10x +\frac{350}{14} = -\frac{19}{14}\)

\(x^2 - 10x + 25 = -\frac{19}{14}\)

\((x-5)^2 = -\frac{19}{14}\)

\(x-5 = \sqrt{\frac{19}{14}}i\)

\(x = 5 + \sqrt{\frac{19}{14}}i\)

\(y = 10 - (5 + \sqrt{\frac{19}{14}}i)\)

\(y = 5 - \sqrt{\frac{19}{14}}i\)

 

Therefore, one ordered pair of complex numbers that works is 5 + sqrt(19/14)i and 5 - sqrt(19/14)i

 

Please let me know if I had made any errors!

 Jun 10, 2024
edited by Maxematics  Jun 10, 2024

1 Online Users

avatar