Question 235575
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Let *[tex \Large x] represent one of the numbers.  Let *[tex \Large y] represent the other number.


Their sum is 20:  (I use the present tense because if their sum was 20, their sum certainly still is 20 -- unless one or both of the numbers have changed in which case the problem cannot be solved with the information given)


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 20]


or equivalently:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 20\ -\ x]


The sum of their squares is still (hopefully) 300:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ y^2\ =\ 300]


But having an expression for *[tex \Large y] in terms of *[tex \Large x], we can make a convenient substitution:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ (20\ -\ x)^2\ =\ 300]


Expand the squared binomial and collect like terms to get:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^2\ -\ 40x\ +\ 100\ =\ 0]


Now all that is left for you to do is to 1. Verify that last step.  2. Solve the quadratic for *[tex \Large x].  3.  Subtract *[tex \Large x] from 20 to determine *[tex \Large y]. And finally, 4. Multiply *[tex \Large xy]


Note:  You will need to either complete the square or use the quadratic formula to solve the quadratic. Leave the roots in radical form.  As it will turn out, if you have solved the quadratic correctly, the sum of the two roots will be 20 -- hence one of the roots is the value of *[tex \Large x] and the other is the value of *[tex \Large y].  Remembering the factorization of the difference of two squares will be helpful performing the last step.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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