How do you find the zeros, real and imaginary, of #y= -2x^2-x-19 # using the quadratic formula?

Question

Abigail Allen · Accepted Answer

Zeros of #f(x)# are #{-1/4+isqrt151/4,-1/4-isqrt151/4}# To find zeros of #f(x)=-2x^2-x-19#, we need roots of the equation #-2x^2-x-19=0#, which can be obtained using quadratic formula. As the roots of #ax^2+bx+c=0# are #x=(-b+-sqrt(b^2-4ac))/(2a)# Hence, roots of zeros of #-2x^2-x-19=0# are #x=(-(-1)+-sqrt((-1)^2-4(-2)(-19)))/(2(-2))# = #(1+-sqrt(1-152))/(-4)=-1/4+-isqrt151/4# Zeros of #f(x)# are #{-1/4+isqrt151/4,-1/4-isqrt151/4}#

Eva Abramson · Answer

undefined To find the zeros of the quadratic equation y = -2x^2 - x - 19 using the quadratic formula, we first identify the coefficients a, b, and c. Then, we substitute these values into the quadratic formula:

Quadratic formula: $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $

For the equation $ y = -2x^2 - x - 19 $:
a = -2
b = -1
c = -19

Substituting these values into the quadratic formula:

$ x = \frac{{-(-1) \pm \sqrt{{(-1)^2 - 4(-2)(-19)}}}}{{2(-2)}} $

Simplify:

$ x = \frac{{1 \pm \sqrt{{1 - 152}}}}{{-4}} $

$ x = \frac{{1 \pm \sqrt{{-151}}}}{{-4}} $

Since the discriminant (b^2 - 4ac) is negative, the solutions will be imaginary. Therefore, the zeros of the equation are complex numbers.