Question

If $x+y=3$ and $xy=2$, then find $x^4+y^4$.

Answer 1

We have,

$x+y=3.......\left(1\right)$

$xy=2.......\left(2\right)$

By equation (2) and we get,

$y=\frac{2}{x}$

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

$x+\frac{2}{x}=3$

$x^2+2=3x$

$x^2-3x+2=0$

$x^2-(2+1)x+2=0$

$x^2-2x-1x+2=0$

$x(x-2)-1(x-2)=0$

$(x-2)(x-1)=0$

If $x-1=0$ then, $x=1$

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

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

$y=\frac{2}{1}=2$

$y=\frac{2}{2}=1$

Therefore,

$x^4+y^4=2^4+1^4=17$

Hence, this is the answer.