Question

how many real zeros does polynomial x2+x+1 have.


PLEASE HELP SOLVE IT

Answer 1

Solution of quadratic equation(ax2+bx+c=0)ax2+bx+c=0)is given by -

x=−b±b2−4ac√2ax=−b±b2−4ac2a. . . (1)

Here, a = 1, b = 1 and c = 1 (x2+x+1=0)(x2+x+1=0)

Just put the value of a, b and c in eq(1),

x=−1±12−4∗1∗1√2∗1x=−1±12−4∗1∗12∗1

x=−1±−3√2x=−1±−32

This is shows its have only imaginary roots.

As i=−1−−−√.i=−1.

x=−1±i3√2x=−1±i32

So two roots are,

x=−1+i3√2x=−1+i32 and x=−1−i3√2.

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Answer 2

Answer:

No real zeros.

Step-by-step explanation:

For checking no. of zeros lets calculate value of D or Discriminant.

$b^{2} - 4ac$

Our equation is x2+x+1.

here, a = 1, b = 1 and c = 1

Lets put these values in D.

$1^{2} - 4(1)(1)$ = -3

We know that if value of D for a quadratic equation is less than 0 then there are no real zeros of that polynomial.

Hope this helps.