In order to solve for the zeros, we equate the given equation to zero.
Let y=0
3x^2-6x+9=03x2−6x+9=0
divide both sides of the equation by 3
x^2-2x+3=0x2−2x+3=0
We now take note of the numerical coefficients of each term.
1*x^2+(-2)*x+3=01⋅x2+(−2)⋅x+3=0
The quadratic equation is given by
ax^2+bx+c=0ax2+bx+c=0
We can see clearly that a=1a=1 and b=-2b=−2 and c=3c=3
The quadratic formula for solving the unknown x is given by
x=(-b+-sqrt(b^2-4*a*c))/(2a)x=−b±√b2−4⋅a⋅c2a
direct substitution
x=(-(-2)+-sqrt((-2)^2-4*1*3))/(2*1)x=−(−2)±√(−2)2−4⋅1⋅32⋅1
x=(2+-sqrt(4-12))/2x=2±√4−122
x=(2+-sqrt(-8))/2x=2±√−82
x=(2+-2sqrt2i)/2x=2±2√2i2
x=1+-sqrt2 ix=1±√2i
The zeros are
x_1=1+sqrt2 ix1=1+√2i
x_2=1-sqrt2 ix2=1−√2i
Check at x_1=1+sqrt2 ix1=1+√2i using the original equation
3x^2-6x+9=03x2−6x+9=0
3(1+sqrt2 i)^2-6(1+sqrt2 i)+9=03(1+√2i)2−6(1+√2i)+9=0
3(1+2sqrt2i+(sqrt2)^2*i^2)-6-6sqrt2 i+9=03(1+2√2i+(√2)2⋅i2)−6−6√2i+9=0
3(1+2sqrt2i+(2)*(-1))-6-6sqrt2 i+9=03(1+2√2i+(2)⋅(−1))−6−6√2i+9=0
3(1+2sqrt2i-2)-6-6sqrt2 i+9=03(1+2√2i−2)−6−6√2i+9=0
3(2sqrt2i-1)-6-6sqrt2 i+9=03(2√2i−1)−6−6√2i+9=0
6sqrt2i-3-6-6sqrt2 i+9=06√2i−3−6−6√2i+9=0
0=00=0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check at x_1=1-sqrt2 ix1=1−√2i using the original equation
3x^2-6x+9=03x2−6x+9=0
3(1-sqrt2 i)^2-6(1-sqrt2 i)+9=03(1−√2i)2−6(1−√2i)+9=0
3(1-2sqrt2i+(-sqrt2)^2*i^2)-6+6sqrt2 i+9=03(1−2√2i+(−√2)2⋅i2)−6+6√2i+9=0
3(1-2sqrt2i+(2)*(-1))-6+6sqrt2 i+9=03(1−2√2i+(2)⋅(−1))−6+6√2i+9=0
3(1-2sqrt2i-2)-6+6sqrt2 i+9=03(1−2√2i−2)−6+6√2i+9=0
3(-2sqrt2i-1)-6+6sqrt2 i+9=03(−2√2i−1)−6+6√2i+9=0
-6sqrt2i-3-6+6sqrt2 i+9=0−6√2i−3−6+6√2i+9=0
0=00=0
God bless...I hope the explanation is useful.
