Verified answer

3x²-6x-1= 0

3x²-6x-1= m² - 2mn + n²

m² = 3x² => m = V3x

2mn = 6x => 2V3xn = 6x => n = 3/V3

So:

(V3x - 3/V3)² = 3x² - 6x + 3

(V3x - 3/V3)² - 4 = 3x² - 6x + 3 - 4

(V3x - 3/V3)² - 4 = 3x² - 6x -1

(V3x - 3/V3)² - 4 = 0

(V3x - 3/V3)² = 4

V3x - 3/V3 = -2 or V3x - 3/V3 = 2

Let x' and x" the roots . So:

V3x' - 3/V3 = -2

V3x' = -2 + 3/V3

V3x' = (-2V3 + 3)/V3

x' = -2/V3 + 1

V3x" - 3/V3 = 2

V3x" = 2 + 3/V3

x" = 2/V3 + 1

3x²-6x-1 = a(x-x')(x-x")

3x²-6x-1 = 3(x- (-2/V3 + 1))(x-(2/V3 + 1)) (answer 1)

3x²-6x-1 = 3(x + 2/V3 - 1)(x- 2/V3 - 1)

3x²-6x-1 = 3( (x -1)+ 2/V3) ((x- 1) - 2/V3)

3x²-6x-1 = 3[(x -1)² - (2/V3)²]

3x²-6x-1 = 3[(x -1)² - (4/3)]

3x²-6x-1 = 3(x -1)² - 4 (answer 2)

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About the

x²-2x+3; a=1, b = -2, c = 3

Note that x²-2x+3 haven´t real roots:

delta = b² - 4ac = 4 - 4*1*3 = 4 - 12 = -8

Let x' and x" be the roots of x²-2x+3

x' = (-b-Vdelta)/2a = (2-2V2i)/2 = 2(1-V2i)/2 = (1-V2i)

x" = (-b+Vdelta)/2a = (2+2V2i)/2 = 2(1+V2i)/2 = (1+V2i)

x²-2x+3 = (x-x')(x-x")

x²-2x+3 = (x-(1-V2i))(x-(1+V2i)) (answer 1)

x²-2x+3 = (x-1+V2i)(x-1-V2i)

x²-2x+3 = ((x-1)+V2i)((x-1)-V2i)

x²-2x+3 = (x-1)² - (V2i)²

x²-2x+3 = (x-1)² - (2i²)

x²-2x+3 = (x-1)² - (2*-1)

x²-2x+3 = (x-1)² + 2 (answer 2)

x²-2x+3 = (x²-2x+1) + 2

Kisses from Brazil

=**

Firstly, you didn't divide the equation by 3 correctly. But are you sure you're supposed to divide the equation? Maybe you're supposed to complete the square using the original equation.

3x² - 6x - 1 = 0

3x² - 6x = 1

3(x² - 2x) = 1

Completing the square:

The coefficient of the x term is -2.

Half of -2 is -1.

Square the -1 and add to the polynomial:

3(x²-2x + (-1)²)

To keep the equation balanced you have to add THREE TIMES (-1)² to the right side.

3(x²-2x + (-1)²) = 1 + 3(-1)²

3(x-1)² = 4

(x-1)² = 4/3

x-1 = ±√(4/3)

= ±2/√3

x = 1 ± 2/√3

= 1 ± (2√3)/3

3x²-6x-1=0

3x²-6x-1+4=0+4

3x²-6x+3=4

3(x²-2x+1)=4

(x²-2x+1)=4/3

(x-1)²=4/3

(x-1) = ±√(4/3)

x=1+√(4/3) and

x= 1-√(4/3)

i'm assuming dat u want us to solve for x--rite??

Therefore the process of completing is:

(3x²-6x+9=10)/3

well u r a bit off here:

it is

3x²-6x+3=4

now

(x-1)² = 4/3

so x= 1 + 2/rt(3) and 1 - 2/rt(3)

here rt--> square root!

3x²-6x-1=0

3x^2-6x=1

divinding by 3

x^2-2x=1/3

x^2-2x+1=1/3+1

(x-1)^2=4/3

x-1=+-root(4/3)

x=((2 sqrt(3))/(3))+1

x=-((2 sqrt(3))/(3))+1

FACTORS ARE (X+1)(X-3)

y r u using completing the square method...??

use another..simpler than this...!