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
We want to use elimination to solve
y = 0 (1)
x + y = 40 (2)
Multiply (1) by -1 to eliminate y.
-y = 0 (3)
Add (2) and (3).
x + y + (-y) = 40 + 0
x = 40
Answer: Multiply by -1.
<h3><u>given</u><u>:</u></h3>
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<h3><u>solution</u><u>:</u></h3>
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1) (3-x) x (3+x)
2) 5(5-x) x (5+x)
3) (3v-wy) x (3v+wy)
4) 2(k-m) x (k+m) x (k^2+m^2)
5) (ab-4c) x (ab+4c)
Brainliest?
