X^2+x+1=0 What is the answer?

Answer 1

Please be specific on what needs to be done. I assume your solving for "x", correct?

First: since the expression can't be factored - use the Quadratic Equation which is....

x = [- b +/- V`(b^2 - 4ac)] / 2a

Sec: you have three coefficients for three variables...
a = 1
b = 1
c = 1

*Replace the coefficients with the corresponding variables....

x = [- 1 +/- V`(1^2 - 4(1)(1))] / 2(1)

x = [- 1 +/- V`(1^2 - 4(1)(1))] / 2

x = [- 1 +/- V`(1^2 - 4)] / 2

x = [- 1 +/- V`(1 - 4)] / 2

x = [- 1 +/- V`(- 3)] / 2

*Rule - you can't find the square root of a negative number. The negative sign becomes an imaginary number, represented as the "i" variable.

x = [- 1 +/- i V`3] / 2

Third: you have two solutions - one is positive, the other is negative.....

a. x = [- 1 +/- i V`3] / 2
x = - 1/2 + (i V`3)/2

b. x = [- 1 - i V`3] / 2
x = - 1/2 - (i V`3)/2

Solutions: - 1/2 + (i V`3)/2 and - 1/2 - (i V`3)/2

Answer 2

normally, for something in the form of Ax^2 + Bx + C (standard form) you would use "reverse FOIL." example: we're given
x^2 + x - 6
and asked for an answer. you would:
1. take all possible combinations of the C variable, -6.
-1*6 or 1*-6
-2*3 or 2*-3
2. now see which ones can add or subtract into B. we notice
-1 + 6 = 5
1 - 6 = -5
-2 + 3 = 1
2 - 3 = -1
-2 and 3 give us 1, which is the B variable. at this point, input your variables into
(x + n1)(x + n2)

to get
(x+3) (x-2)
now, to get these = to 0, you must make the #s in the ()'s equal 0. watch, i'll show you how to find a solution:
((-3) + 3)((-3) - 2)
by plugging in -3 for the x i get
(0)(-5)
(0)
the -3 is a solution. but where did i get it? simple: -3 is the only number that will cancel out 3 and get 0 (it's "inverse"). so, we need 1 more solution... guess what it is....

yep, the inverse of 2, which is -2. problem solved! but in your case, there's a hidden 1:
x^2 + x + 1 = 0
x^2 +1x +1 = 0
anything times one is the number, so you could say there's a one there. now factor it:
x^2 + 1x + 1
1*1
1+1 = 2
1 and 1 is the omny possible combination, yet it won't work. but you have an ace up your sleeve: the good old quadratic formula!
_______________
-b +- -/ b^2 - 4 (ac)
____________________
2a
heh heh, nice radiacl, eh? Answers needs a math-friendly system for questions like yours. anyways... using standard form...
1x^2 + 1x + 1
________________
-1 + -/ 1^2 - 4 (1) (1)
_____________________
2(1)
____
-1 +- / 1-4
---------------
2
now, 1-4 is -3, meaning that there is no answer. Negative numbers have no square roots.

well, there you go! much ado about nothing!

Answer 3

idk what grade lvl u r,so this is algebra 2 lvl

first factor (X+1) (X+1)
make 2 eq. X+1=0 X+1=0
move the 1's X+1-1=0-1 X+1-1=0-1
combine same terms X=-1 X=-1
combine equal answers X=-1
*important rule- X does not equal under 0 & X does not equal 1
no solution

Answer 4

It would not ingredient actual so i will finished the sq.: x^2 + x + a million = 0 x^2 + x + (a million/2)^2 + a million = (a million/2)^2 (x + a million/2)^2 + a million = a million/4 (x + a million/2)^2 = -3/4 x + a million/2 = 0.866i or x + a million/2 = -0.866i x = 0.866i - a million/2 or x = -0.866i - a million/2

Answer 5

Using the quadratic formula:
a = 1, b = 1, c = 1

x = [-1 ± √(1² - 4*1*1)]/2*1
x = [-1 ± √(1 - 4)]/2
x = (-1 ± √-3)/2

Answer 6

Use the quadratic formula:
x=-b±√b^2-4ac/2a
a=1,b=1,c=1
x=-1±√(1)^2-4(1*1)/2(1)
x=(-1±√1-4)/2
x=(-1±√-3)/2

There is no real solution unless you use imaginary numbers! Here's an example:
x=(-1±√3i)/2

Answer 7

There is no solution.

If the equation is
y = x^2 + x + 1...
It's asking when y = 0 which is NOWHERE.

This is part of the family of the parabola equation (x^2)
Now that parabola goes through y only once which is at its minimum:
(0,0) this is the y-intercept which there is only one of.

This particular equation states that the y-intercept is (+1).
This moves the graph 1 unit upwards.
IF the entire graph is shifted up 1, then the minumum is now: (0,1)

therefore the y NEVER equals 0 because the minimum is 1.
Therefore at y = 0, x = N.S.

Answer 8

x= [ - 1 ±√(1 - 4) ] / 2

x = [ - 1 ±√(-3)] / 2

Complex number, i , is required for √(-3) where i² = -1

x = [ - 1 ±√(3i²)] / 2

x = [-1 ± √3i ] / 2

It may be however that original equation should have read:-
x² + 2x + 1 = 0
(x + 1)(x + 1) = 0
x = - 1 (with no need for complex numbers!)

Answer 9

-1+V-3/2,-1-√-3/2

Answer 10

Good show, Nissi. You nailed it ☺

The roots are
-(1/2) + i√3/2 and
-(1/2) - i√3/2.

And if hitmast3r doesn't understand what i means or what complex roots are, that's his problem.


Doug