Completing the Square to solve 2x^2 + 8x + 1 = 0

[iPwn]

I'm puzzled on this one:

. . .Solve 2x² + 8x + 1 = 0 by completing the square.

So you subtract 1 to both sides, then divide by 2, getting:

. . .x² + 4x = -1/2

. . .(1/2 * 4)² = 4

. . .(x+2)² = 7/2 = (3 1/2)

. . .x + 2 = +/- √ (7/2)

. . .x = -2 +/- √ (7/2)

However the answer in the book is "x = -2 +/- √(14) / 2'. (The "2" is not inside the radical sign.) So they must have had √14/4 and squared the 4. But WHERE THE HECK did they get 14/4???

Thanks!

 

[skeeter]

x² + 4x = -1/2

x² + 4x + 4 = -1/2 + 4

(x + 2)² = 7/2

x + 2 = +/- √(7/2)

x + 2 = +/- √(7/2) * √(2/2) = [√(7)*√(2)]/[√(2)*√(2)

x + 2 = +/- √(14)/2

x = - 2 +/- √(14)/2

 

[iPwn]

Well how did you multiply by 2/2 on just ONE side???

If you do something to the right side you have to do it to the left correct?

Beyond that multiplying by the square root of 2/2 is 1, so are you saying that the square root of 7/2 is exactly the same as the square root of 14 divided by 2?

If so then why isn't the square root of 7/2 the answer.

 

[Denis]

Well golly gee iPwn, if you're having problems completing the square,
why don't you solve using the quadratic first...then you'll have something
"to aim for" ; get my drift?