3x^2-2x-4=0

Question

How Do I Solve

Answer 1

Karin is correct, above - the quadratic formula can be used to solve this equation. However, it's not the only way to do it. You could also use the "completing the square" method. Although I think the quadratic formula is easier, and it will absolutely work for any quadratic equation you could be given, some homework or test questions may specifically ask for you to solve a quadratic using the "completing the square" method, so I'll show how to use it below:

 

To do so, we will take the typical ax² + bx + c = 0 format, and move the constant "c" term to the other side of the equation:

 

       3x² - 2x -4 = 0

 

       3x² - 2x = 4

 

Next, if the "a" in front of the x² term is anything other than 1, we will divide it out of each term in the equation:

 

       x² - (2/3)x = 4/3

 

Now, the hardest part, we add the square of half of the new "b" constant in front of the x term to both sides of the equation. In this case, ((1/2)*(2/3))² = 1/9, so 1/9 is added to both sides:

 

       x² - (2/3)x + (1/9) = 4/3 + 1/9 = 12/9 + 1/9 = 13/9

 

This process allows us to factor the left side of the equation into:

 

       (x - 1/3)² = 13/9

 

Now we take the square root of both sides, being careful to remember that the square root of any given positive number could be positive or negative. (For example, the square root of 9 is either 3 or -3). In practice, this means that we can set up two different equations, to give us our two possible x-value solutions to this quadratic, one where the constant on the right side of the equation is positive, and one where it is negative:

 

        x - 1/3 = +/- sqrt(13)/3

 

        x = 1/3 + sqrt(13)/3                  OR                  x = 1/3 - sqrt(13)/3

 

 

 

        x = (1 + sqrt(13))                      OR                 x = (1 - sqrt(13))

                      3                                                                     3

Answer 2

3x^2-2x-4=0

use the quadratic equation

x^2+(-2/3)x+(-4/3)

 

{2/3+sqrt[(4/9)+(16/3)}/2

{2/3+sqrt[4/9+48/9]}/2

{2/3+sqrt[52/9]}/2

{2/3+sqrt[13*(4/9)]}/2

{2/3+(2/3)sqrt(13)}/2

multiply both terms by 1/2

1/3+(1/3)sqrt(13)

1/3 + (1/3)sqrt(13)

1/3 - (1/3)sqrt(13)

or

[1+sqrt(13)]/3

[1-sqrt(13)]/3

Answer 3

you must use the quadratic formula bc there are no factors of -12 that will sum to a -2. So 

 

X=(-b + or - sqrt(b² - 4ac))/2a

(- -2 + or - sqrt((-2)² -4 (3)(-4)))/2(3))

(2 + or - sqrt(4 +48))/6

(2 + or - sqrt(52))/6

(2 + or - 2 Sqrt(13))/6

factor out a 2 

2(1 + or - sqrt(13))/6

(1+ sqrt (13))/3 and (1-sqrt(13))/3