X^2+1=0
What are the roots of this equation?
Thanks!

X² + 1 = 0

=> (x+1)² - 2x = 0

=> x+1 = √(2x)

or x - √(2x) + 1 = 0

Now take y=√x

So, the equation changes to

y² - y√2 + 1 = 0

By quadratic formula, we get:-

y = [√2 ± √(2–4)]/2

or √x = (√2 ± i√2)/2 or (1 ± i)/√2 [by cancelling the √2 in numerator and denominator and ‘i' is a imaginary number with value √(-1)]

or x = [(1 ± i)²]/2

So roots are [(1+i)²]/2 and [(1 - i)²]/2

Thus we got two roots but in complex plane. If you put this values in the formula for formation of quadratic equation, that is x²+(a+b)x - ab where a and b are roots of the equation, you will get the equation

x² + 1 = 0 back again
So it’s x=1 or x=-1