Answer to Question #153393 in Algebra for Bee

Question

x²+1=0

y=x²+1

find the solution of the given system of equations.

Answer 1

$\pmb{x^2+1=0}\;\;\; ..(1)\\ \pmb{y=x^2+1}\;\;\;..(2)$

Now from equation $(1),$

$x^2+1=0\\ \Rightarrow (x)^2-(-1)=0\\ \Rightarrow (x)^2-(i)^2=0\\ \pmb[where \;\;i=\sqrt{-1}\;\pmb]\\ \Rightarrow (x-i)(x+i)=0\\ \Rightarrow x=i\;\;\;or, \;\;\;x=-i$

Now putting these $x$-values in equation $(2)$ ,

$y=0$ if $x=i$

and, $y=0$ if $x=-i$

$\therefore$ From above discussion, we can decide that,

Solution set of $(x,y)=\{(i,0),(-i,0)\}$ [Ans.]

i.e. $(x,y)= (i,0) \;\;\;or, \;\;\;(-i,0)$