Question

help.........................​

Answer 1

x+y = 1

xy( xy -2) = 12

$x {}^{4} + y {}^{4} \\ \\( x {}^{2} + y {}^{2} ) {}^{2} - 2x {}^{2} y {}^{2} \\ \\ ((x + y) {}^{2} - 2xy) {}^{2} - 2x {}^{2} y {}^{2} \\ \\ (x + y) {}^{4} + 4x {}^{2} y {}^{2} - 4(x + y) {}^{2} xy - 2x {}^{2} y {}^{2} \\ \\ (x + y) {}^{4} - 4(x + y) {}^{2} xy + 2x {}^{2} y {}^{2} \\ \\ as \: x + y = 1 \: and \: xy(xy - 2) = 12 \\ \\ x {}^{2} y {}^{2} - 2xy = 12 \\ \\ (x + y) {}^{4} - 4(x + y) {}^{2}xy + 2x {}^{2} y {}^{2} \\ \\ 1 - 4xy + 2x {}^{2} y {}^{2} \\ \\ 1 + 2( - 2xy + x {}^{2} y {}^{2} ) \\ \\ = 1 + 2(12) \\ \\ = 1 + 24 \\ \\ = 25$

Answer 2

Given,

$x+y=1\ \ \&\ \ xy(xy-2)=12$

We have to find,

$x^4+y^4.$

First let me take the product given. Means, let me find the value of xy first.

$xy(xy-2)=12$

This product can be written as,

$(xy-1+1)(xy-1-1)=12$

Here, let  $xy-1=k.$

So the equation becomes,

$(k+1)(k-1)=12$

On expanding the LHS, we get,

$k^2-1=12\ \ \Longrightarrow\ \ k^2=13\ \ \Longrightarrow\ \ k=\sqrt{13}$

Taking the value of k that we considered,

$xy-1=\sqrt{13}\ \ \Longrightarrow\ \ xy=1+\sqrt{13}$

We got xy.  Also, we need  (xy)².

$(xy)^2=(1+\sqrt{13})^2=14+2\sqrt{13}$

Now, let's consider the given sum.

$x+y=1$

We square both sides of this equation.

$(x+y)^2=1^2\ \ \Longrightarrow\ \ x^2+y^2+2xy=1$

Since  $\mathsf{xy=1+\sqrt{13}\ \ \Longrightarrow\ \ 2xy=2+2\sqrt{13}\ ,}$  we subtract the corresponding sides from our equation. I.e.,

$x^2+y^2+2xy-2xy=1-(2+2\sqrt{13})\\ \\ \Longrightarrow\ x^2+y^2=-1-2\sqrt{13}$

Repeat the same procedure again by squaring both sides again.

$(x^2+y^2)^2=(-1-2\sqrt{13})^2\ \ \Longrightarrow\ \ x^4+y^4+2(xy)^2=53+4\sqrt{13}$

And we have  $(xy)^2=14+2\sqrt{13}\ \ \Longrightarrow\ \ 2(xy)^2=28+4\sqrt{13}.$  As we did ago,  we subtract corresponding sides from our equation. I.e.,

$x^4+y^4+2(xy)^2-2(xy)^2=53+4\sqrt{13}-(28+4\sqrt{13})\\ \\ \Longrightarrow\ x^4+y^4=53+4\sqrt{13}-28-4\sqrt{13}=\bold{25}$

Thus the answer is 25.