How do you find the zeros, real and imaginary, of #y=3x^2-6x+9# using the quadratic formula?

1 Answer

Answer:

the zeros are
#x_1=1+sqrt2 i#
#x_2=1-sqrt2 i#

Explanation:

In order to solve for the zeros, we equate the given equation to zero.

Let y=0
#3x^2-6x+9=0#
divide both sides of the equation by 3
#x^2-2x+3=0#
We now take note of the numerical coefficients of each term.

#1*x^2+(-2)*x+3=0#

The quadratic equation is given by
#ax^2+bx+c=0#
We can see clearly that #a=1# and #b=-2# and #c=3#

The quadratic formula for solving the unknown x is given by

#x=(-b+-sqrt(b^2-4*a*c))/(2a)#

direct substitution

#x=(-(-2)+-sqrt((-2)^2-4*1*3))/(2*1)#

#x=(2+-sqrt(4-12))/2#

#x=(2+-sqrt(-8))/2#

#x=(2+-2sqrt2i)/2#

#x=1+-sqrt2 i#

The zeros are
#x_1=1+sqrt2 i#
#x_2=1-sqrt2 i#

Check at #x_1=1+sqrt2 i# using the original equation

#3x^2-6x+9=0#
#3(1+sqrt2 i)^2-6(1+sqrt2 i)+9=0#
#3(1+2sqrt2i+(sqrt2)^2*i^2)-6-6sqrt2 i+9=0#
#3(1+2sqrt2i+(2)*(-1))-6-6sqrt2 i+9=0#

#3(1+2sqrt2i-2)-6-6sqrt2 i+9=0#

#3(2sqrt2i-1)-6-6sqrt2 i+9=0#

#6sqrt2i-3-6-6sqrt2 i+9=0#

#0=0#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check at #x_1=1-sqrt2 i# using the original equation

#3x^2-6x+9=0#
#3(1-sqrt2 i)^2-6(1-sqrt2 i)+9=0#
#3(1-2sqrt2i+(-sqrt2)^2*i^2)-6+6sqrt2 i+9=0#
#3(1-2sqrt2i+(2)*(-1))-6+6sqrt2 i+9=0#

#3(1-2sqrt2i-2)-6+6sqrt2 i+9=0#

#3(-2sqrt2i-1)-6+6sqrt2 i+9=0#

#-6sqrt2i-3-6+6sqrt2 i+9=0#

#0=0#

God bless...I hope the explanation is useful.