Question

If x + y = 1 and xy(xy-2) = 12, then the
value of x4 + y4 is:​

Answer 1

Answer:

Please see the attachment

Answer 2

$Given \: x + y = 1 \: ---(1) \\and \: xy(xy-2) = 12 \: ---(2)$

/* We know that, */

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

$= 1 - 2xy \: [From \: (1) ] \: ---(3)$

/* On squaring equation (3) , we get */

$Now , (x^{2} + y^{2})^{2} = (1-2xy)^{2}$

$\implies (x^{2})^{2} + (y^{2})^{2} + 2x^{2}y^{2} = 1^{2} + (2xy)^{2} - 2 \times 1 \times 2xy$

$\implies x^{4} + y^{4} + 2x^{2}y^{2} = 1+4x^{2}y^{2} - 4xy$

$\implies x^{4} + y^{4} = - 2x^{2}y^{2} + 1+4x^{2}y^{2} - 4xy$

$\implies x^{4} + y^{4} = 1+2x^{2}y^{2} - 4xy$

$\implies x^{4} + y^{4} = 1+2\times xy( xy- 2)$

$\implies x^{4} + y^{4} = 1+2\times 12\: [From \:(2) ]$

$\implies x^{4} + y^{4} = 25$

Therefore.,

$\red {Value \: of \: x^{4} + y^{4}}\green { = 25}$

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