CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Number of real ordered pair $$(x, y)$$ satisfying $$x^{2}+1=y$$ and $$y^{2}+1=x$$ is


A
0
loader
B
1
loader
C
2
loader
D
4
loader

Solution

The correct option is A $$0$$
$$x^{2}+1=y$$ and $$y^{2}+1=x$$
Subtracting $$x^{2}-y^{2}=y-x$$
$$\Rightarrow \left ( x-y \right )\left ( x+y+1 \right )=0$$
$$\Rightarrow  x=y $$ or $$  x+y+1  =0$$
If $$x=y$$ then $$x^{2}+1=y \Rightarrow x^{2}-x+1=0$$ which has no real roots.
If $$ x + y + 1 = 0$$ then $$ x + y = -1$$
Adding given equations, $$x^{2}+y^{2}+2=x+y$$
$$\Rightarrow x^{2}+y^{2}+2=-1 \Rightarrow x^{2}+y^{2}+3=0$$ which is not possible.

Mathematics

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image