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4-2.Quadratic Equations and Inequations
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The number of ordered pairs $(x, y)$ of real numbers that satisfy the simultaneous equations $x+y^2=x^2+y=12$ is
A
$0$
B
$1$
C
$2$
D
$4$
(KVPY-2015)
Solution
(d)
We have,
$x \quad x+y^2=x^2+y =12$
$x+y^2 =12$
$x^2+y =12 \ldots \text {…(ii) }$
On subtracting Eq.$(i)$ from Eq. $(ii)$, we get
$\left(x^2-x\right)+\left(y-y^2\right)=0$
$\Rightarrow \quad\left(x^2-y^2\right)-(x-y)=0$
$\Rightarrow \quad(x-y)(x+y-1)=0$
$\Rightarrow \quad x=y, x+y=1$
When $x=y$
$\therefore \quad x^2+x=12 \Rightarrow x^2+x-12=0$
$\Rightarrow(x+4)(x-3)=0$
$x=-4,3, y=-4,3$
When $x+y=1$
$\therefore \quad x^2+1-x=12$
$\Rightarrow x^2-x-11=0$
$\Rightarrow \quad x=\frac{1 \pm \sqrt{45}}{2}$
$\therefore$ Total number of ordered pair of $(x, y)$ is $4 .$
Standard 11
Mathematics
