(x^2+y^2)^2=(X^2-y^2)^2+(2xy)^2 determine the sum of the squares of two numbers if the difference of the squares of the numbers is 5 and te product of the number is 6

Guest Apr 17, 2020

#1**0 **

\((x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2\)

The problem gives us that:

\(x^2-y^2 =5\)

and

\(xy =6\)

Realize that we already have all the necessary values to determine \(x^2+y^2 \) without individually solving for x or y

Substituting the values of \(x^2-y^2\) and \(xy\) that we already have:

\((x^2+y^2)^2 = (5)^2 + (2*6)^2\)

\((x^2+y^2)^2=25+(12)^2= 25 + 144 = 169\)

Now we square root both sides to get:

\(x^2+y^2 = \pm13\). The problem doesn't directly limit x and y to real numbers, so we could have an answer of -13(square rooting 169 gives a negative and positive solution). The implied answer however would be **13**.

jfan17 Apr 17, 2020