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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 $833$.
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