2007-11-23 08:07:09 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Solve x^2 + 2x - 1 = 0 first for boundaries,
x = -1±√2
Since x^2 + 2x - 1 < 0 opens up, we have
-1-√2 < x < -1+√2
2007-11-23 08:24:31 · answer #1 · answered by sahsjing 7 · 0⤊ 0⤋
The area between the graph of y = x^2 +2x - 1 and the x-axis (for y<0).
What was the question?
2007-11-23 08:23:44 · answer #2 · answered by Polly 1 · 0⤊ 0⤋
x^2 + 2x - 1 < 0?
let it = k
x^2 + 2x - 1 = k
or x^2 + 2x - (1+ k) =0
Then test the result for values of K that give < 0
2007-11-23 08:20:35 · answer #3 · answered by Anonymous · 0⤊ 0⤋
x^2 + 2x - 1 < 0
complete the square
x^2 + 2x + 1 < 2
(x+1)^2 < 2
x+1 < sqrt2
x < sqrt2 - 1 or about .4
also x > -1 - sqrt2 or about -2.4
(sqrt2 -1) > x > (-sqrt2 - 1)
2007-11-23 08:22:47 · answer #4 · answered by Steve A 7 · 0⤊ 0⤋
-1 plus or minus the square root of 2 is less than 0 use a calulator to figure out whic one works
2007-11-23 08:15:18 · answer #5 · answered by HI!!!!!! 3 · 0⤊ 1⤋
the answer is
-2 - sqrt2 < x < -2 + sqrt2
2007-11-23 14:13:28 · answer #6 · answered by Aslan 6 · 0⤊ 0⤋
Yes, sometimes.
What that the question you meant to ask?
2007-11-23 08:14:55 · answer #7 · answered by Tom V 6 · 0⤊ 1⤋