How do you find the zeros, real and imaginary, of #y= -9x^2-28x-73# using the quadratic formula?

Question

Isaiah Anthony · Accepted Answer

Zeros of #y=-9x^2-28x-73# are #-14/9-sqrt(461)/9i# and #-14/9+sqrt(461)/9i# To find zeros of #y=-9x^2-28x-73#, one needs to find values of #x# for which #y=f(x)=0#. For #ax^2+bx+c=0#, solution is given by quadratic formula #(-b+-sqrt(b^2-4ac))/(2a)#. Hence for #-9x^2-28x-73=0# or #9x^2+28x+73=0#, #x=(-28+-sqrt(28^2-4xx9xx73))/(2xx9)# = #(-28+-sqrt(784-2628))/18# = #(-28+-sqrt(-1844))/18# = #(-28+-sqrt(-4xx461))/18# = #-14/9+-sqrt(461)/9i#

Eva Abramson · Answer

undefined To find the zeros of the quadratic function $y = -9x^2 - 28x - 73$ using the quadratic formula:

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

where $a = -9$, $b = -28$, and $c = -73$.

Substitute these values into the quadratic formula:

$$x = \frac{{-(-28) \pm \sqrt{{(-28)^2 - 4(-9)(-73)}}}}{{2(-9)}}$$

Simplify:

$$x = \frac{{28 \pm \sqrt{{784 - 2628}}}}{{-18}}$$

$$x = \frac{{28 \pm \sqrt{{-1844}}}}{{-18}}$$

Since the discriminant ($b^2 - 4ac$) is negative, the solutions will be complex. We can simplify the expression under the square root by factoring out $-4$ from $1844$ and extracting the square root of $-1$:

$$x = \frac{{28 \pm \sqrt{{-4 \cdot 461}}}}{{-18}}$$

$$x = \frac{{28 \pm 2i\sqrt{461}}}{{-18}}$$

Therefore, the zeros of the function are complex numbers:

$$x = \frac{{28 \pm 2i\sqrt{461}}}{{-18}}$$