Quadratic formula?

Question

3x^2-2x+1=0

Answer 1

x = [2 +/- sqrt(4-4*3*1)]/2*3

= (2 +/- sqrt(-8))/6 = (2 +/- 2i*sqrt2)/6 = (1 +/- i*sqrt2)/3

Answer 2

the quadratic formula is x= (-b +/-(b^2-4ac)^(1/2))/2a.....(1)

for the quadratic equation ax^2 +bx+c=0

here,a=3,b= -2 and c=1

substitute into (1) we have

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

= (2 +/-(4-12)^(1/2)/6 = ( 2+/-(-8)^(1/2))/6

= (2 +/-2(2)^(1/2)i)/6

=( 1 +/-(2)^(1/2)i)/3 where i = (-1)^(1/2)

x is equal to 1 plus or minus root 2 times i divided by 3 ,where i is the complex number root(-1)

in this answer root means the square root

i hope this helps

Answer 3

To solve ax^2+bx+c=0:

x=(-b +/- sqrt(b^2-4ac)) / (2a)

In this case, a=3, b=-2, and c=1, so

x=(2 +/- sqrt(4-12)) / 6 = 1/3 +/- i*sqrt(2)/3,

where i = sqrt(-1), the imaginary unit.

Answer 4

This has no solutions. It meas the graph would never equals 0 (never crosses x axis).

You can check by doing b^2-4ac. Its negative so has no solutions. When you do the sq root in the formula you cant do the sq root of a neg number.

Answer 5

Compare the given equation 3x^2-2x+1=0
with the standard quadratic equation ax^2+bx+c=0
Here a=3, b=-2 and c=1
x=[-(b) +/-{sq rt(b^2-4ac)}]/2a
By substituting values of a,b,c we get
x=[-(-2) +/-{sq rt(-2^2-4*3*1)}]/2*3
x=[2 +/-{sq rt(4-12)}]/6
x=[2 +/-{sq rt(-8)}]/6
x=[2 +/-{2sq rt(-2)}]/6
x=[1 +/-{sq rt(-2)}]/3

Answer 6

You just plug the numbers in, something like negative 2a plus or minus the aha here you go http://www.purplemath.com/modules/quadform.htm so just pull out the coefficients and plug those in.