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# Im stuck...

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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.

Mar 21, 2020

#1
+402
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(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)

Mar 21, 2020
#2
+1612
+2

Nice, jfan17!!!

CalTheGreat  Mar 21, 2020
#3
+22085
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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
#4
+111
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$$\displaystyle x^{4}=y^{4}-176\sqrt{7},\\ \text{so}\\ 176\sqrt{7}=y^{4}-x^{4}=(y^{2}+x^{2})(y^{2}-x^{2})=22(y+x)(y-x)=88(y-x),\\ \text{so} \\ y-x=2\sqrt{7}, \\ x-y = -2\sqrt{7}.$$

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Mar 21, 2020
#5
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Thank you so so soo much!

Guest Mar 21, 2020