imaginary numbers and completing the square

Question

2x^2-8x=-1 seems unfactorable. I used the quadratic formula and got 2+or- .5sqrrtof14. Since this is positive, does it have any imaginary numbers or solutions for that matter? Also I am confused as to how to complete the square.

 

Thanks

Answer 1

2x² - 8x = -1

 

When completing the square, we want one side of the equation to look something like this:  (x ± c)² , where c is a constant.

 

First, let's divide both sides of the equation by 2.  This will allow the x² term on the left side of equation to have a coefficient of 1.

 

x² - 4x = -1/2

 

 

Next, add 4 on both sides of the equation to the left side of equation so that we get a perfect square.

 

x² - 4x + 4 = (-1/2) + 4

 

(x - 2)(x - 2) = 7 / 2

 

(x - 2)² = 7 / 2

 

 

Square both sides of the equation to undo the square-root.

 

x - 2 = ±√(7 / 2)

 

 

Add 2 on both sides of the equation.

 

x = 2 ± √(7 / 2)

 

x = 2 ± (√7 / √2)

 

x = [2√2 ± √7] / √2

 

 

Rationalize the denominator.  Multiply the numerator and denominator by √2.

 

x = (4 ± √14) / 2

 

x = 2 ± (1/2)√14

 

 

The solution when completing the square matches with the one when you used the quadratic formula.

 

Answer 2

2x² - 8x = -1  Divide both sides by 2 to get a 1 for the lead coefficient

 

x² - 4x = -1/2  Complete the square by adding (1/2*4)² to both sides

 

x² - 4x + 4 = -1/2 + 4  Factor and simplify

 

(x-2)² = 7/2  Solve for x

 

x-2 = +/- sqrt(7/2)

 

x = +/- Sqrt(7/2) +2   Rationalize the denominator and combine the terms

 

x - +/- [Sqrt(14) +4]/2