Find in simplest radical form the roots of x square-6x+6=0?

Answer 1

for any quardratic equation of following form

aX^2 + bX + C = 0

two roots of X would be

[- b +/- sqrt (b^2 - 4ac)]
--------------------------------
2a

using this formula the two roots of the equation you gave would be

6 +/- sqrt (36-24)
----------------------
2

or x1 = [6 +2 sqrt(3)]/2 = 3 + sqrt (3)
x2 = 3 - sqrt (3)

in numbers it will be (sqrt 3 = 1.732)

x1= 4.732; x2 = 1.268

cool

Answer 2

Since this doesn't factor cleanly, you can either use the quadratic equation or complete the square. The quadratic is easy if you can remember it, but I prefer completing the square because it's intuitive and easy to remember:

1. Put the constant on the right side of the equation:
x^2 - 6x = -6

2. Take half the coefficient of x (in this case, -6), square it, and add it to both sides:
x^2 - 6x + 9 = 3

3. Factor the left side to (x + a)^2, where a is the number you just squared:
(x - 3)^2 = 3

4. Take the square root of both sides, remembering that this will leave you both a positive and a negative root:
x + 3 = +/- sqrt 3

5. Isolate x by moving the constant on the left to the right side:
x = -3 +/- sqrt 3

Answer 3

You need to use the quadratic equation, and hence A is 1, B is -6, and C is 6.

This yields this in the Quadratic Equations:

[6+-(sqre root)(12)]/2

The square root of 12 can be reduced to 2*the square root of 3:

6+2*(Sqre root)(3)/2 and

6-2*(Sqre root)(3)/2

Reduced:

3+(sqre root) (3) and 3-(sqre root) (3)

Answer 4

x^2 - 6x + 6 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-6) ± sqrt((-6)^2 - 4(1)(6)))/(2(1))
x = (6 ± sqrt(36 - 24))/2
x = (6 ± sqrt(12))/2
x = (6 ± sqrt(4 * 3))/2
x = (6 ± 2sqrt(3))/2
x = 3 + sqrt(3) or 3 - sqrt(3)

Answer 5

x=[6+/-sqrRoot(36-4(1)(6)]/2
= [6+/-root(12)]/2
=3+/-root(3)

Answer 6

hmmmmmmmmmm
x^2 -5x+6 works, but not -6x in the middle

^^^^^^^^^^^^^^
would be (x-3)(x-2)= x squared -5x + 6

or if it is -7x

(x-6)(x-1) = x^2 -7x +6