solve by completing the square
x^2+6x-1=0
2007-04-16 14:14:49 · 5 answers · asked by gingin 1 in Science & Mathematics ➔ Mathematics
x² + 6x - 1 = 0
now x² + 6x are the first two terms of the expansion of
(x+3)² = x² + 6x + 9
Hence x² + 6x is the same as (x+3)² - 9 and we have
(x+3)² - 9 - 1 = 0
(x+3)² - 10 = 0
(x+3)² = 10
Take √ of both sides and do not forget the ±!
x + 3 = ± √10
x = -3 ± √10
Hope this helps.
2007-04-16 14:18:22 · answer #1 · answered by M 6 · 4⤊ 1⤋
x^2 + 6x - 1 = 0
x^2 + 6x - 1 +10 = 10
x^2+ 6x + 9 = 10
(x+3)^2 = 10
x+3 = +/- (10)^(1/2)
x = -3 +/- (10)^(1/2)
2007-04-16 21:20:02 · answer #2 · answered by Brian F 4 · 0⤊ 1⤋
x^2+6x-1=0
x^2+6x+9-9-1=0
(x+3)^2=1
x+3=1 or -1
x=-2 or -4
2007-04-16 21:20:28 · answer #3 · answered by iyiogrenci 6 · 0⤊ 1⤋
You need to factor out the equation to two seperate parentheses that got multiplied together. For example:
(x+2)(x+4) = x^2+6x+8
Now you have to backwards to see what the two factors were.
2007-04-16 21:20:17 · answer #4 · answered by SEOP 1 · 0⤊ 1⤋
x = -6 +- sqrt. 36 +4 over 2
x = -6 +- sqrt 40 over 2
x = -6 +- 2sqrt. 10 over 2
x = 3 +-sqrt 10
2007-04-16 21:27:35 · answer #5 · answered by Anonymous · 0⤊ 0⤋