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4-2.Quadratic Equations and Inequations
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The number of pairs of reals $(x, y)$ such that $x=x^2+y^2$ and $y=2 x y$ is
A
$4$
B
$3$
C
$2$
D
$1$
(KVPY-2009)
Solution
(a)
We have,
$x =x^2+y^2$
$\text { and } y =2 x y$
$\because y-2 x y =0$
$\Rightarrow y(1-2 x) =0$
$\Rightarrow y=0, x =\frac{1}{2}$
Put $y=0$ in Eq. (i), we get
$x =x^2+0$
$\Rightarrow x-x^2=0$
$\Rightarrow x(1-x)=0$
$\Rightarrow x=0, x=1$
$\text { Put } x=\frac{1}{2} \text { in Eq. (i), we get }$
${\frac{1}{2}=\left(\frac{1}{2}\right)^2+y^2}$
$\Rightarrow y^2=\frac{1}{2}-\frac{1}{4}$
$\Rightarrow \quad y^2=\frac{1}{4} \Rightarrow y=\pm \frac{1}{2}$
$\therefore \text { Value of }(x, y) \operatorname{are}(0,0)(0,1)$
$\left(\frac{1}{2}, \frac{1}{2}\right)\left(\frac{1}{2}, \frac{-1}{2}\right)$
Standard 11
Mathematics
