# If $x^2 + y^2 = 29$ and $xy = 2$, find the value of(i) $x + y$(ii) $x - y$(iii) $x^4 + y^4$

Given:

$x^2 + y^2 = 29$ and $xy = 2$

To do:

We have to find the value of

(i) $x + y$

(ii) $x - y$

(iii) $x^4 + y^4$

Solution:

The given expressions are $x^2 + y^2 = 29$ and $xy = 2$. Here, we have to find the value of (i) $x + y$ (ii) $x - y$ (iii) $x^4 + y^4$. So, by using the identities $(a+b)^2=a^2+2ab+b^2$ and $(a-b)^2=a^2-2ab+b^2$, we can find the required values.

$xy = 2$.........(I)

$(a+b)^2=a^2+2ab+b^2$.............(II)

$(a-b)^2=a^2-2ab+b^2$.............(III)

(i)

Let us consider,

$x^2 + y^2 = 29$

Adding $2xy$ on both sides, we get,

$x^2+2xy+y^2=29+2xy$

$(x+y)^2=29+2(2)$                    [Using (II) and (I)]

$(x+y)^2=29+4$

$(x+y)^2=33$

Taking square root on both sides, we get,

$(x+y)=\pm\sqrt{33}$

The value of $(x+y)$ is $\pm\sqrt{33}$.

(ii)

Let us consider,

$x^2 + y^2 = 29$

Subtracting $2xy$ on both sides, we get,

$x^2-2xy+y^2=29-2xy$

$(x-y)^2=29-2(2)$                    [Using (III) and (I)]

$(x-y)^2=29-4$

$(x-y)^2=25$

Taking square root on both sides, we get,

$(x-y)=\sqrt{25}$

$x-y=\pm 5$

The value of $x-y$ is $\pm 5$.

(iii)

Let us consider,

$x^2 + y^2 = 29$

Squaring on both sides, we get,

$(x^2 + y^2)^2 = (29)^2$

$x^4+2x^2y^2+y^4=841$

$x^4+2(xy)^2+y^4=841$                        (Since $a^mb^m=(ab)^m$)

$x^4+2(2)^2+y^4=841$                          [Using (I)]

$x^4+2(4)+y^4=841$

$x^4+8+y^4=841$

$x^4+y^4=841-8$                       (Transposing $8$ to RHS)

$x^4+y^4=833$

The value of $x^4+y^4$ is $833$.