The real numbers x and y satisfy

\(\begin{align*} x + y &= 4, \\ x^2 + y^2 &= 22, \\ x^4 &= y^4 - 176 \sqrt{7}. \end{align*}\)

Compute x - y.

Note: I calculated \(4-2\sqrt{7}\), and \(2\sqrt{7}-4, -\sqrt{7}+2\), but they didn't work.

Guest Mar 21, 2020

#1**+1 **

(x+y)^2 = 16

x^2+2xy+y^2 = 16

x^2 + y^2 = 22

2xy = -6

xy = -3

y = -3/x

(x - 3/x)^2 = 16

x^2 - 6 + 9/(x)^2 = 16

x^2 + 9 / (x)^2 = 22

Completing the square, we can get:

x^2 + 6 + 9/(x)^2 = 28

This is useful, because it asks us for x-y. Remembering our earlier substitution of y = -3/x:

x - y = x - (-3/x) = x + 3/x

(x+3/x)^2 = 28

(x+3/x) = **sqrt(28)**

jfan17 Mar 21, 2020

#3**0 **

I found similar BUT

x+y = 4 is given .... then if

x-y = sqrt28 then if you add two equations together

2x = 4 + sqrt28 = 4 + 2 sqrt 7

then

x = 2 + sqrt 7

Now x+y = 4 means 2+ sqrt7 + y = 4 then y = 4-(2 + sqrt7) = 2 - sqrt7

Then does x^2 + y^2 = 22 ??? Yes it does

Does x^4 = y^4 - 176 sqrt7 ? No it does not,,,,, I do not think the question is written correctly ! Did you find this too?

ElectricPavlov
Mar 21, 2020