x^2-x+8=0
2007-03-29 06:12:53 · 7 answers · asked by Meowmix 2 in Science & Mathematics ➔ Mathematics
If you want to find the roots, there ain't any real roots, cos The discriminant,
D= b^2 - (4*a*c) is negative., ie -31
but complex roots would be
[1 + i Sqrt(31)]/2
[1 - i Sqrt(31)]/2
where i = sqrt(-1)
2007-03-29 06:18:27 · answer #1 · answered by shrek 5 · 0⤊ 1⤋
You need the quadratic formula. Any equation of the form ax^2+bx+c=0 has a solution
x=(-b+sqrt(b^2-4ac))/(2a) and another solution x=(-b-sqrt(b^2-4ac))/(2a)
This means that there are two real solutions whenever b^2>4ac, one real solution whenever b^2=4ac, and two complex solutions (involving imaginary numbers) whenever b^2<4ac.
This is an example of the last class, so the only solutions are complex numbers.
2007-03-29 06:19:11 · answer #2 · answered by jiyuztex 2 · 0⤊ 2⤋
x=+4, x=-4
2007-03-29 06:19:38 · answer #3 · answered by Eric K 1 · 0⤊ 1⤋
Use the quadratic formula. a=1, b=-1 and c=8. Expect imaginary roots.
2007-03-29 06:22:20 · answer #4 · answered by cattbarf 7 · 0⤊ 2⤋
quadradic equation- (-b+/- the square root of (B^2-4AC)) divided by 2A....you get (-1+/-the square root of -31) divided by 2.....so you get two answers because its plus or minus (+/-)
2007-03-29 06:23:38 · answer #5 · answered by tlfdfirefighter 2 · 0⤊ 1⤋
(1 +or- sqrt (-31))/2
2007-03-29 06:18:19 · answer #6 · answered by Maverick 7 · 0⤊ 2⤋
This parabola has no x- zeros, its y- intercept is (0,8)
Vertex (.5,7.75)
2007-03-29 06:29:22 · answer #7 · answered by dwinbaycity 5 · 0⤊ 0⤋