Given-that-x-y-R-x-2-y-2-32-x-y-4-x-y-4-4352-Find-the-value-of-x-2-y-2-

Question Number 115555 by ZiYangLee last updated on 26/Sep/20

Given that x, y \in R \forall x^{2} - y^{2} = 32,

{(x + y)}^{4} + {(x - y)}^{4} = 4352,

Find the value of x^{2} + y^{2} .

Commented by Rasheed.Sindhi last updated on 27/Sep/20

A n O ther W ay

(B y c h a n g i n g i n t o

r e c i p r o c a l f o r m)

{(x + y)}^{4} + {(x - y)}^{4} = 4352

\frac{{(x + y)}^{4}}{{(x + y)}^{2} {(x - y)}^{2}} + \frac{{(x - y)}^{4}}{{(x + y)}^{2} {(x - y)}^{2}}

= \frac{4352}{{(x + y)}^{2} {(x - y)}^{2}}

\frac{{(x + y)}^{2}}{{(x - y)}^{2}} + \frac{{(x - y)}^{2}}{{(x + y)}^{2}} = \frac{4352}{{(x^{2} - y^{2})}^{2}}

\frac{{(x + y)}^{2}}{{(x - y)}^{2}} + \frac{{(x - y)}^{2}}{{(x + y)}^{2}} = \frac{4352}{{(32)}^{2}} = \frac{17}{4}

\frac{{(x + y)}^{2}}{{(x - y)}^{2}} = z

z + \frac{1}{z} = \frac{17}{4}

4 z^{2} - 17 z + 4 = 0

(z - 4) (4 z - 1) = 0

z = 4 \lor z = \frac{1}{4}

\frac{{(x + y)}^{2}}{{(x - y)}^{2}} = 4, \frac{1}{4}

\frac{{(x + y)}^{2}}{{(32 / (x + y))}^{2}} = 4, \frac{1}{4}

{(x + y)}^{4} = 4 {(32)}^{2}

{(x + y)}^{2} = \pm 64

{(x - y)}^{2} = {(32 / (x + y))}^{2} = 32^{2} / (\pm 64) = \pm 16

2 (x^{2} + y^{2}) = {(x + y)}^{2} + {(x - y)}^{2}

x^{2} + y^{2} = \frac{1}{2} (\pm 64 \pm 16) = \pm 40

Answered by mr W last updated on 26/Sep/20

u = x + y

v = x - y

x^{2} - y^{2} = (x + y) (x - y) = 32

\Rightarrow u v = 32

u^{4} + v^{4} = 4352

u^{4} + v^{4} + 2 u^{2} v^{2} = 4352 + 2 {(u v)}^{2}

{(u^{2} + v^{2})}^{2} = 4352 + 2 \times 32^{2} = 6400 = 80^{2}

\Rightarrow u^{2} + v^{2} = 80

u^{2} + v^{2} + 2 u v = 80 + 2 (u v)

{(u + v)}^{2} = 80 + 2 \times 32 = 144 = 12^{2}

\Rightarrow u + v = \pm 12

u^{2} + v^{2} - 2 u v = 80 - 2 (u v)

{(u - v)}^{2} = 80 - 2 \times 32 = 16 = 4^{2}

\Rightarrow u - v = \pm 4

x = \frac{u + v}{2} = \pm 6

y = \frac{u - v}{2} = \pm 2

\Rightarrow x^{2} + y^{2} = 36 + 4 = 40

Commented by ZiYangLee last updated on 26/Sep/20

Thanks very much . .

Answered by ruwedkabeh last updated on 26/Sep/20

{(x + y)}^{4} + {(x - y)}^{4} = 4352,

x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 {x y}^{3} + y^{4} + x^{4} - 4 x^{3} y + 6 x^{2} y^{2} - 4 {x y}^{3} + y^{4} = 4352

2 x^{4} + 12 x^{2} y^{2} + 2 y^{4} = 4352

x^{4} + 6 x^{2} y^{2} + y^{4} = 2176

{(x^{2} - y^{2})}^{2} + 8 x^{2} y^{2} = 2176

{(32)}^{2} + 8 x^{2} y^{2} = 2176

1024 + 8 x^{2} y^{2} = 2176

8 x^{2} y^{2} = 1152

x^{2} y^{2} = 144

{(x^{2} + y^{2})}^{2} = {(x^{2} - y^{2})}^{2} + 4 x^{2} y^{2} = 1024 + 576 = 1600

x^{2} + y^{2} = 40

Commented by ZiYangLee last updated on 26/Sep/20

Thank you!

Answered by Olaf last updated on 26/Sep/20

{(x + y)}^{4} + {(x - y)}^{4} =

{[{(x + y)}^{2} + {(x - y)}^{2}]}^{2} - 2 {(x + y)}^{2} {(x - y)}^{2} =

{(2 x^{2} + 2 y^{2})}^{2} - 2 {(x^{2} - y^{2})}^{2} =

4 {(x^{2} + y^{2})}^{2} - 2 \times 32^{2} = 4352

{(x^{2} + y^{2})}^{2} = \frac{4352 + 2048}{4} = \frac{6400}{4} = 1600

x^{2} + y^{2} = 40

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