Question

If x and y are two numbers (real or imaginary) such that (x + y)4 + (x – y)4 = 1600 and x2 – y2 = 80. Find the value of (x2 + y2)

Answer 1

$Given \: x^{2} - y^{2} = 80 \: ---(1) \\and \: (x+y)^{4} + (x-y)^{4} = 1600 \: --(2)$

$\implies [(x+y)^{2}]^{2} + [(x-y)^{2}]^{2} = 1600$

$\implies [ (x+y)^{2} + (x-y)^{2}]^{2} - 2 (x+y)^{2} (x-y)^{2} = 1600$

$\boxed{\pink{ \because a^{2}+b^{2} = (a+b)^{2} - 2ab }}$

$\implies [ 2(x^{2}+y^{2})^{2} ]^{2} - 2[(x+y)(x-y)]^{2} = 1600$

$\implies 4(x^{2}+y^{2})^{2}- 2(x^{2} - y^{2})^{2} = 1600$

$\implies 4(x^{2}+y^{2})^{2}- 2\times 80^{2} = 1600\: [ From \: (1) ]$

$\implies 4(x^{2}+y^{2})^{2}- 12800 = 1600$

/* Dividing each term by 4, we get */

$\implies (x^{2}+y^{2})^{2}- 3200 = 400$

$\implies (x^{2}+y^{2})^{2} = 3200 +400$

$\implies (x^{2}+y^{2})^{2} = 3600$

$\implies x^{2} + y^{2} = \pm \sqrt{60^{2}}$

$\implies x^{2} + y^{2} = \pm 60$

Therefore.,

$\red{ Value \:of \: x^{2} + y^{2}}\green { = \pm 60}$

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