Question

If $x^2+y^2=25,xy=12,$ then $x$ is equal to:

• A. $3,4$
• B. $3,-3$
• C. $3,4,-3,-4$
• D. $-3,-3$

Answer 1

Answer: (C)

We have, $x^2+y^2=25,xy=12$
$\Rightarrow x^2+{\left(\frac{12}{x}\right)}^2=25$
$\Rightarrow x^4-25x^2+144=0$
$\Rightarrow x^4-16x^2-9x^2+144=0$
$\Rightarrow \left(x^2-16\right)\left(x^2-9\right)=0$
$\Rightarrow x=\pm 4$ and $x=\pm 3$
Alternate Solution:
We have, $x^2+y^2=25$ and $xy=12$
Now, $x^2+y^2+2xy=25+2(12)$
$\Rightarrow (x+y{)}^2=49$
$\Rightarrow x+y=\pm 7\dots (i)$
and $x^2+y^2-2xy=25-2(12)$
$\Rightarrow (x-y{)}^2=1$
$\Rightarrow x-y=\pm 1$
On solving Eqs. (i) and (ii), we get
$x=\pm 4$ and $x=\pm 3$