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:
Step-by-step explanation:
125 = 5 *5 * 5 = 5³
8 = 2 * 2 *2 = 2³
125x⁶ - 8 = 5³(x²)³ - 2³
= (5x²)³ - 2³ { a³ - b³ = (a -b)(a² + ab + b²)
= (5x² - 2) ([5x²]² + 5x²*2 + 2²)
= (5x² - 2)(25x⁴ + 10x² + 4)
Hint: (5x²)² = 5² * (x²)² = 25* x²ˣ² = 25x⁴
Answer:
B) 5y ( x-2 )
Step-by-step explanation:
= 5xy -10y
= 5y ( x-2 )
Answer:
10^3
Step-by-step explanation:
15-12=3
10^3
