How do you find the zeros, real and imaginary, of #y=-x^2-4x-11# using the quadratic formula?

Question

Eva Abramson · Accepted Answer

2 imaginary roots:
#x = - 2 +- isqrt7# #y = - x^2 - 4x - 11 = 0# Use the improved quadratic formula in graphic form (Google Search) #D = d^2 = b^2 - 4ac = 16 - 44 = - 28# Since D < 0, there are 2 imaginary roots. #d = +- 2isqrt7# #x = -b/(2a) +- d/(2a) = 4/-2 +- 2isqrt7/2 = - 2 +- isqrt7#

Avery Alexander · Answer

undefined To find the zeros of the quadratic equation $ y = -x^2 - 4x - 11 $ using the quadratic formula:

1. Identify the coefficients: $ a = -1 $, $ b = -4 $, and $ c = -11 $.
2. Substitute these coefficients into the quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $.
3. Plug in the values: $ x = \frac{{-(-4) \pm \sqrt{{(-4)^2 - 4(-1)(-11)}}}}{{2(-1)}} $.
4. Simplify inside the square root: $ x = \frac{{4 \pm \sqrt{{16 - 44}}}}{{-2}} $.
5. Further simplify: $ x = \frac{{4 \pm \sqrt{{-28}}}}{{-2}} $.
6. Since the square root of a negative number results in an imaginary number, the zeros will be complex.
7. Simplify the square root of -28: $ \sqrt{{-28}} = \sqrt{{-1 \times 28}} = \sqrt{{-1}} \times \sqrt{{28}} = i \sqrt{{28}} = 2i\sqrt{{7}} $.
8. Substitute this value into the equation: $ x = \frac{{4 \pm 2i\sqrt{{7}}}}{{-2}} $.
9. Simplify: $ x = -2 \pm i\sqrt{{7}} $.
10. So, the zeros are $ x = -2 + i\sqrt{{7}} $ and $ x = -2 - i\sqrt{{7}} $.