Question

number of real roots of x^2+x+1=0

Answer 1

Answer:

The number of real roots of $x^2+x+1=0$ is 0.

Step-by-step explanation:

A quadratic equation of the form $ax^2+bx+c=0$, has solutions, given by

$x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}$

Also, if the discriminant

$d=b^2-4ac$

is negative, then the equation has complex roots. That is the equation has no real roots.

Given the quadratic equation,

$x^2+x+1=0$

So the coefficients are

$a= 1\\\\b=1\\\\c=1$

Thus the discriminant is,

$d=1^2-4\times 1\times 1=1-4=-3$

Since this quantity is negative, the equation has no real roots.