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