Solving by Completeing the Square?

Question

i suck at math big time!
Ok, I got about halfway, but got stuck, and I now am doubting myself big time. does anyone know how to solve this?

2x^2-8x-19=0

thanks

Answer 1

Move the 19 over & divide by 2

x²-4x = 19/2

The "square" then is (x-2)². Add 4 to both sides

x²-4x + 4 = 19/2 + 8/2

(x-2)² = 27/2
x-2 = ± 3√(3/2)
x = 2 ± 3√(3/2)

Answer 2

The easiest way to solve this is by quadratic formula. If you want to solve by completing the square, here's how:

First, the leading coeffecient must equal 1. Currently, it equals 2. Thus, you must divide by 2:

x^2 -4x - (19/2) = 0

Then, so it won't get in the way, get the constant on the other side:

x^2 - 4x = (19/2)

To complete the square on the left-hand side, you take the coeffecient of the x term (4), divide by 2 (2), and then square it (4). Add this number to both sides of equation:

x^2 -4x +4 = (19/2) + 4

Now, the left-hand side can be factored as a perfect square:

(x - 2) ^2 = (19/2) + (8/2)

(x-2) ^ 2 = (27/2)

Square root both sides:

x - 2 = +-sqrt (27/2)

Solve for x:

x = 2 +-sqrt (27/2)

Simplify according to instructor's directions. For instance,

sqrt (27) = 3*sqrt(3)

You may also need to rationalize the denominator. In the end you should have:

x = 2 +- [3sqrt(6) / 2]

Hope this helps.

Answer 3

x² - 4x - 19/2 = 0
(x² - 4x + 4) - 4 - 19/2 = 0
(x - 2)² - 8/2 - 19/2 = 0
(x - 2)² - 27/2 = 0
(x - 2)² = 27/2
x - 2 = + or - √(27/2 ) = + or - (3√3 / √2)
x - 2 = + or - (3√3√2 / 2) = + or - 3√6 / 2
x = (2 + 3√6 / 2 ), or x = (2 - 3√6 / 2)

Answer 4

2x^2 - 8x - 19 = 0
Dividing both sides by 2
x^2 - 4x - 19/2 = 0
(x)^2 - 2 . x . 2 + (2)^2 - (2)^2 - 19/2 = 0
(x - 2)^2 = (2)^2 + 19/2 = 4 + 19/2 = 27/2
Taking square root of both sides
x - 2 = +- sqrt(27/2)
Adding 2 to both sides
x = 2 + sqrt(27/2) and x = 2 - sqrt(27/2)
Use your calculator and sail

Answer 5

vidigod has the most correct and complete answer.