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
Answer:
A
Step-by-step explanation:
Answer:
6
Step-by-step explanation: You can set up the equation: 
= 
You square both sides to cancel out the squares which leaves you with 
= n - 4. Then you just solve as normal which ends you up with an answer of 6. You can check this by plugging 6 back into the equation and see that you end up with 
on both sides.
