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# help

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(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

Apr 17, 2020

#1
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$$(x^2+y^2)^2=(x^2-y^2)^2+(2xy)^2$$

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.

Apr 17, 2020