solve for x?

Question

4x^2 - 6x + 2 = 0
show your work
so i can understand please

Answer 1

4x^2 - 6x + 2 = 0

Divide the equation by two:

2x^2 - 3x + 1 = 0

Factor:

(2x - 1)(x - 1) = 0

x = 1/2, x = 1

Answer 2

A quadratic equation with real (or complex) coefficients has two (not necessarily distinct) solutions, called roots, which may be real or complex, given by the quadratic formula:


2ax= - b- ( b^2-4ac)^.5, and 2a x= -b+(b^2-4ac)^.5
____________
There fore x= (6-2) / 8 = 1/2 and x = (6+2)/8 = 1

Checking by substitution in original 4x^2 - 6x + 2 = 0

For x = 1/2, 4/4 - 3 + 2 = 0 OK

For x=1 4(1) - 6 + 2 + 0 OK

Answer 3

4x^2 - 6x + 2 = 0
4x^2 - 4x - 2x + 2 = 0
4x(x - 1) - 2(x - 1) = 0
(4x - 2)(x - 1) = 0

x = 1, x = 1/2

Suppose (ax^2 + bx + c) is a quadratic equation. To factor it, split b in two numbers that when multiplied gives you the product of a and c.

Answer 4

4x^2 - 6x + 2

factor out 2
2 (2x^2 - 3x + 1) = 0

factor. Find two numbers that multiply to 2 and add to -3

the two numbers are -1 and -2. Since the leading coefficent is not 1, replace the middle terms

2 (2x^2 - 2x - 1x + 1) = 0

factor by groups
2 [ (2x^2 - 2x) + (-1x + 1) ] = 0

factor out GCF
2 [ 2x (x - 1) + -1(x - 1) ] = 0

2 [ 2x (x - 1) - 1(x - 1) ] = 0

factor out (x - 1)
2 (x - 1) (2x - 1) = 0

x = 1 or 1/2

Answer 5

4x²- 6x + 2 = 0

4x² - 4x - 2x + 2 = 0

Group factor

(4x² - 4x) - (2x + 2) = 0

4x(x - 1) - 2(x - 1) = 0

(4x - 2)(x - 1) = 0

- - - - - - - - -s-

Answer 6

I think this is how you do it
first you factor
and then you'll get
(2x-1)(2x+2) = 0
then you find out what numbers subsitiuted for X will give you zero in each parentheses
so
x= -1,1/2

Answer 7

4x^2 - 6x + 2 = 0 // divide by 4

x^2 - 1.5x + 0.5 = 0

Completing the square:
x^2 - 2*0.75x + 0.75^2 - 0.75^2 + 0.5 = 0

(x - 0.75)^2 - 0.5625 + 0.5 = 0

(x - 0.75)^2 - 0.0625 = 0

(x - 0.75)^2 - 0.25^2 = 0

(x - 0.75 - 0.25)(x - 0.75 + 0.25) = 0

(x - 1)(x - 0.5) = 0

x1 = 1
x2 = 0.5