x^2 + 5x - 14 = o
if there is more than one root, separate them with commas
2006-11-04 07:57:56 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2 + 5x - 14 = 0
x^2 -2x + 7x -14 = 0
x(x -2) + 7(x-2) = 0
(x + 7)(x - 2) = 0
Either x+7 or x-2 = 0
Put x + 7 = 0
x= -7
Put x-2 = 0
x = 2
There are two roots
x = 2,-7
2006-11-05 03:06:32 · answer #1 · answered by Akilesh - Internet Undertaker 7 · 0⤊ 0⤋
If you write your equation as ax^2 + bx + c, the roots by the quadratic equation are [-b +/- sqrt(b^2 - 2ac)] / 2a. In your equation, a = 1, b = 5, and c = -14, so the formula simplifies to [-5 +/- sqrt(25 + 14)] / 2, which I'm sure you can calculate for yourself. You'll have two roots, because b^2 - 2ac != 0.
2006-11-04 08:02:18 · answer #2 · answered by DavidK93 7 · 0⤊ 0⤋
Factor
(x+7)(x-2)=0
set each factor =0
x+7=0 or x-2=0
solve each
x=-7 or x=2
2006-11-04 08:36:09 · answer #3 · answered by mom 7 · 0⤊ 0⤋
x^2 + 5x - 14 = 0
(x-2)(x+7)=0
x-2=0
x=2
x+7=0
x=-7
x=2, -7
2006-11-04 08:07:04 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
-7 and 2. graph it, or go to his site: http://www.shodor.org/interactivate/activities/GraphSketcher/
2006-11-04 08:10:35 · answer #5 · answered by Anonymous · 0⤊ 0⤋
-7,2
2006-11-04 08:28:15 · answer #6 · answered by yourdaddy0 2 · 0⤊ 0⤋