# If z and z' are the roots of the equation x^2-x+1=0 prove that z+z'=1 and z*z'=1

hala718 | Certified Educator

x^2 - x + 1 = 0

z and z' are roots for the equation.

a= 1      b =-1      c = 1

But we know from Viete's rule,

z + z' = -b/a = -(-1/1) = 1

==> z + z' = 1

Also ,

we know that:

z*z' = c/a =1/1 = 1

==> z*z' = 1

giorgiana1976 | Student

If z and z' are the roots of the equation x^2-x+1=0, then if we'll substitute them into equation, they will verify it.

We'll use Viete's relations to express the sum and the product of the roots of the equation:

z + z' = -b/a

z*z' = c/a

We'll identify a,b,c.

a = 1

b = -1

c = 1

z + z' = 1/1

z + z' = 1

z*z' = 1/1

z*z' = 1

We could also prove that, by applying the quadratic formula:

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

z = (1 + i*sqrt3)/2

z' = (1 - i*sqrt3)/2

Now, we'll calculate the sum z + z':

z + z' = (1 + i*sqrt3+1 - i*sqrt3)/2

We'll eliminate like terms:

z + z' = 2/2

z + z' = 1

We'll calculate the product z*z':

z*z' = (1 + i*sqrt3)*(1 - i*sqrt3)/4

z*z' = (1 - 3*i^2)/4 , where i^2 = -1

z*z' = (1+4)/4

z*z' = 4/4

z*z' = 1

neela | Student

The equation x^2-x+1 = 0. z and z' are the roots of the equation

To show that z+z' = 1 and z*z' = -1.

Solution:

The assumed roots be z and z'.

By the relation between the roots and coefficients 9in the case of a quadratic) equation,

The sum of the roots = z+z' = - coefficient of x / coeeficient of

x^2  = - (-1) =1

The product of roots  z*z'= constant term / coefficient of x^2

= 1/1 = 1.

So z+z' = 1

and zz' = 1.