x^2 - 6x + 9 = -1 + 9. We need to solve for x.
First, add one on each side of the equation:
x^2 - 6x + 9 + 1 = -1 + 9 + 1
x^2 - 6x + 10 = 9
Then subtract 9 from each side:
x^2 - 6x + 10 - 9 = 9-9
x^2 - 6x + 1 = 0
Now we got an equation = 0. This an equation from the second degree, it represented by a parabola which turns up since a>0.
This equation is the developed form as ax^2 + bx + c with a=1; b = -6; and c=1.
Now, to find the zeroes of this equation, we need to find delta Δ.
Δ = b^2 - 4ac.
If Δ>0, the equation admits 2 zeroes: x=(-b-√Δ)/2a and x = (-b+√Δ)/2a.
If Δ<0, the equation doesn't admits any zero.
If Δ=0, the equation admits one zero x = -b/2a
Δ = (-6)^2 - 4(1)(1)
Δ = 36 - 4
Δ = 32
Δ>0
So the zeroes of the equation are:
x = (-b-√Δ)/2a = (6-4√2)/2
x = (-b+√Δ)/2a = (6+4√2)/2
Hope this Helps! :)
