2007-09-01 10:32:45 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
5 (x ² + 2 x - 2 / 5)
5 [ (x ² + 2 x + 1) - 1 - 2 / 5 ]
5 [ (x + 1) ² - 7 / 5 ]
5 (x + 1) ² - 7
2007-09-05 08:43:03 · answer #1 · answered by Como 7 · 0⤊ 0⤋
5x^2 + 10x -2
5x^2 + 10x = 2
x^2 + 2x = (2/5)
use (b/2)^2 = (2/2)^2= 1
x^2 + 2x + __________ = (2/5) +__________
x^2 + 2x + 1= (2/5) + 1
x^2 + 2x +1 = 1 2/5
x^2 + 2x + 1 = 7/5
(x +1)^2= 7/5 <== take the square out
(x+1) = square root of 7/5
x = square root of 7/5 substract 1
*remember to put the "postive and negative sign" before the square root
2007-09-01 17:56:03 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I assume you mean solve for x:
5x^2 +10x-2 =0
x^2 + 2x = 2/5
x^2+2x +1 = 1 2/5
(x+1)^2 = 7/2
x+1 = +/- sqrt(7/2)
x = -1 +/- sqrt(3.5)
2007-09-01 17:43:14 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
5x²+10x-2 =0
1/2b^2 = 1
5(x²+2x+1) =2 + 5
5(x²+2x+1) =7
5((x+1)(x+1)) =7
5((x+1)²) =7
5((x+1)²)-7 = 0
2007-09-01 17:50:30 · answer #4 · answered by roOt 3 · 0⤊ 0⤋
5x^2 + 10x - 2 = 5*(x^2 +2x) - 2 = 5(x^2 + 2x + 1) - 2 - 5 =
5*(x + 1)^2 -7 =
[sqrt(5)*(x + 1) + sqrt(7)]*[sqrt(5)*(x + 1) - sqrt(7)].
2007-09-01 17:45:06 · answer #5 · answered by Tony 7 · 0⤊ 0⤋