Question

x^2-14x-4=0, solve by completing the square

Answer 1

here is a lesson on completing the square that might help

http://www.algebra.com/my/Completing_the_Square_001.lesson?content_action=show_dev

Answer 2

x^2-14x-4=0


x^2-14x = 4


x^2-14x + 49 = 4 + 49 ... take half of the x coefficient (-14) to get -7 and square this to get 49. Add this to both sides



(x-7)^2 = 53


(x-7)^2 = 53


x - 7 = +-sqrt(53)


x - 7 = sqrt(53) or x - 7 = -sqrt(53)


x = 7 + sqrt(53) or x = 7 - sqrt(53)


So the solutions are x = 7 + sqrt(53) or x = 7 - sqrt(53)

Answer 3

x^2-14+4=0
to complete the square, we need c (ax^2+bx+c) to equal 49 since b/2 needs to equal c^2

so we will ad 53 to both sides giving us this nice equation:
x^2-14x-4+53=53
x^2-14x+49=53

now we can factor easily
(x-7)(x-7)=53

now put it in standard form:
(x-7)^2 = 53
x-7=+or- squareroot (53)
x=-squareroot(53)+7
and
x=squareroot(53)+7