Question

solve for 0
x^2-x+1=0

Answer 1

i think you mean x?

Answer 2

yes sorry

Answer 3

would it be 0?

Answer 4

use the quadratic formula
x = [-(-1) +/- sqrt(1 - 4 *1 *1) ] / 2

there are no real roots - they are complex

Answer 5

Solve for 0...lol

Answer 6

\[x^3+1=0\]
\[(x+1)(x^2-x+1)=0\]
\[x=-1, x=\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\]

Answer 7

x = 1/2 + i sqrt3/2 and 1/2 - i sqrt3/2

Answer 8

i dont understand any of it

Answer 9

i'm not sure where satellites coming from ? but we agree on the 2 comp;ex roots

Answer 10

haven't you come across complex numbers yet nikkiem?

Answer 11

if not then you should'nt be given this problem - maybe you typed it incorrectly?

Answer 12

yeah but i never really understand them to much sorry im not go at math at all

Answer 13

well - all squares of real numbers are positive 2*2 = 4 , -3 * -3 = 9 etc
A negative number h as no real square root so someone came along and suggested that theres am imaginative number sqrt(-1) which we cal 'i'.

so square root of, say, sqrt( -9 ) = sqrt(9)* sqrt(-1) = = 3i

Answer 14

so my answer then would be -1 right

Answer 15

no the imaginary number in the answer is sqrt (1 - 4) = sqrt (-3) = sqrt3 * sqrt(-1)
= sqrt3 * i or as i wrote it i sqrt3 - i put the i first because it is sqrt 3 not sqrt (3i)
the full answers are 1/2 + i sqrt3 / 2 and 1/2 - i sqrt3 / 2

its best to write them as satellite did using the equation thing but im not good at that.