If $x - y = 3$ and $x y = 10$ , then find $x^{4} + y^{4}$ .

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Solution

We have,

$x - y = 3 . . . . . . (1)$

$x y = 10 . . . . . . (2)$

By equation (2) and we get,

$y = \frac{10}{x}$

Then, put the value of y in equation (1) and we get,

$x - \frac{10}{x} = 3$

$x^{2} - 10 = 3 x$

$x^{2} - 3 x - 10 = 0$

$x^{2} - (5 - 2) x - 10 = 0$

$x^{2} - 5 x + 2 x - 10 = 0$

$x (x - 5) + 2 (x - 5) = 0$

$(x - 5) (x + 2) = 0$

If $x - 5 = 0$ then, $x = 5$

If $x + 2 = 0$ then, $x = - 2$

Put the value of x in equation (2) and we get,

$y = \frac{10}{5} = 2$

$y = \frac{10}{- 2} = - 5$

Value of $x^{4} + y^{4}$

$= {(5)}^{4} + {(2)}^{2}$

$= 625 + 16$

$= 641$

Hence, this is the answer.

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