How do you find the zeros, real and imaginary, of $y= -x^2-2x-1$ using the quadratic formula?

Question

2831 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-26T20:18:16

There are a double rooth, x = - 1.

Applying the Quadratic Formula, we can obtain the rooth of:

$y = - x^2 - 2 x - 1$ ,

solving the equation:

$- x^2 - 2 x - 1 = 0$ .

Remember the Quadratic Formula:

When $a x^2 + b x + c = 0$ , the solutions are:

$x = {- b +- sqrt {b^2 - 4 cdot a cdot c}}/{2 cdot a}$ .

Applying this formula to the equation given, we have:

$x = {- (- 2) +- sqrt {(- 2)^2 - 4 cdot (- 1) cdot (- 1)}}/{2 cdot (- 1)} =$

$= {2 +- sqrt {4 - 4}}/{- 2} = - 1$ .

Because the discriminant is zero, we have only one solution. In this case, the solution is real.

How do you find the zeros, real and imaginary, of y= -x^2-2x-1 y= -x^2-2x-1 using the quadratic formula?