Question

${x}^{2} + 1 = 0$
.
find x. please help




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Answer 1

Step-by-step explanation:

We know that the highest power or order of the equation says the number of solutions that it has. So in this case we must get 2 solutions. But these solutions are in the complex plane.

We can solve the equation as:-

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

The solution that came to my mind was just another way of solving the same equation. If you solve the root [(1+i)²]/2 you will get ‘i’ and if you solve the other root then you will get ‘-i’ which we have been taught for years that x² + 1 = 0 so x = ±√(-1) or ±i

${}^{ \tt \orange{i \: hope \: its \: help \: you}}$