# Complex number!!?

The book states to "solve the equaction in the complex number":

x^2+x+1=0

Relevance

Imaginery number i is as follows:

i^2 = -1 <-----> i = sqrt(-1)

x = [-1 ± √(1-4*1*1)]/2

= [-1 ± √(-3)]/2

= -1/2 ± √(3)*√(-1)/2

= -1/2 ± √(3) /2 * i

• This equation can't be solved using direct factorization

So we use the mathematical formula

x = [-b±√{(b^2) - 4ac}] / 2a

Where a = b = c = 1

Then substitute the values of a, b & c in the formula

You will have 2 answers ( 2 roots for the equations ) which are -1/2 ± (√-3)/2 = -1/2 ± {(√3)/2} i

• Anonymous

Complex number is a fancy name for a number in which sqrt(-1) and its multiples occur.

To solve the above you will have to use the general formula.

=>x1=[(-1) + sqrt(1 - 4)]/2

=>x1=[-1 + sqrt(3)i]/2

=>x2=[(-1) - sqrt(1-4)]/2

=>x2=[-1 -sqrt(3)i]/2

=>x= [-1 + sqrt(3)i]/2 , [-1 -sqrt(3)i]/2

where i=sqrt(-1)

• a = 1 , b = 1 and c = 1

x = [ -b +/- √ ( b^2 - 4ac )]/ 2a

= [ -1 +/- √ ( 1 - 4 )] / 2

= [ -1 +/- √ ( -3 ) ] / 2

= [ -1 +/- √ (3) i ] / 2

x = -1/2 + 1/ 2 √3 i and x = x = -1/2 - 1 / 2 √3 i

• Hmmmm........ And your teacher hasn't said a word about complex numbers, has she (or he)? The words 'imaginary unit' and 'imaginary number' have never been explained to you? They just throw it at you and expect you to figure it all out yourself?

A couple of people here have given very good answers, but are they gonna take your next exam?

Doug

• If you use the quadratic formula, you will find that the two roots are complex numbers. If you graph the equation, you will find that it doesn't cross the X-axis - therefore no real roots.

• x = [- 1 ± √(-3 ) ] / 2

x = [- 1 ± (√3) i ] / 2 is exact answer.

Value of √3 = 1.732 could be used to give an approximate answer.

• x^2 + x + 1 = 0

We apply the quadratic formula, since this cannot be factored:

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

x = (-1 +/- sqrt -3)/2

x = [-1 +/- (i * sqrt 3)]/2

where i = sqrt -1

• complex step. browse onto google and yahoo. just that could help!

• Anonymous