Can someone help me with this algebra problem please?

Question

The question, "One of the solutions of the equation 3x to the 2nd-2x-8=0 is ..." has the answer of -4/3. Can someone tell me how the heck to get this answer?

Answer 1

3x^2-2x-8 = 0
Factor into two binomials.
(3x + 4)(x - 2 ) = 0

Now set each binomial equal to 0 and solve for x.
3x+4=0
3x = -4
x = -4/3

x - 2 = 0
x = 2

Answer 2

3x^2 - 2x - 8 = 0

to determine whether the left-hand side is factorable, determine its discriminant.

B^2 - 4AC = (-2)^2 - 4(3)(-8) = 4 + 96 = 100 = 10^2,

which is most certainly a perfect square; hence, the left-hand side of the original quadratic equation may be factored as follows:

3x^2 - 2x - 8 = 3(x - ((2 + 10)/6))(x - ((2 - 10)/6))
= 3(x - 2)(x + 8/6)
= 3(x - 2)(x + 4/3)
= (x - 2)(3x + 4)

now, to solve for x, set (x - 2) = 0 OR 3x + 4 = 0, then solve these two linear equations separately, to get:

x = 2 OR x = -4/3.

Answer 3

Alternatively use the equation

3x^2 - 2x + 8 = 0

is of the form ax^2 + bx + c = 0

where the roots of the equation are

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

(See the wiki page)

x = (2 +/- SQRT(4 - 4.3.(-8))) / 6

x = (2 +/- SQRT(4 - (-96))/6

x = (2 +/- SQRT(100)) / 6

x = (2 + 10)/6 or (2 -10)/6

therefore

x = 2 or -4/3