How do you find the zeros, real and imaginary, of #y=- x^2+12x+6 # using the quadratic formula?

Question

How do you find the zeros, real and imaginary, of #y=- x^2+12x+6  # using the quadratic formula?

Eva Adamson · Accepted Answer

Solution ; Zeros are two real number #x ~~12.6 , x ~~ -0.48 #

#y=-x^2+12x+6 # Comparing with standard quadratic

equation #ax^2 + bx + c = 0# and #a# is not zero. Here

#a=-1 . b= 12 ,c = 6 ; D# is discriminant and #D=b^2-4ac#

#:.D=b^2-4ac= 12^2-4*(-1)*6= 168# . If #D# is positive,

we get two real solutions, if it is zero just one solution, and

if it is negative we get two complex (imaginary) solutions.

Quadratic Formula: #x = [ -b +- sqrt((b^2-4ac)) ] / (2a)#

#:. x = [ -12 +- sqrt(12^2-4*(-1)* 6)] / ( 2 *(-1)) # or

# x = [ -12 +- sqrt(168)] / (-2) ~~ 6 +- 6.48 # or

#x ~~12.48 (2dp) and x ~~ -0.48 (2dp)#

Solution ; Zeros are two real number #x ~~12.6 , x ~~ -0.48 #

[Ans]

Jameson Allen · Answer

undefined To find the zeros of the quadratic function $y = -x^2 + 12x + 6$ using the quadratic formula, you first identify the coefficients $a$, $b$, and $c$ in the quadratic equation $ax^2 + bx + c = 0$. Then, you substitute these values into the quadratic formula:

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

For $y = -x^2 + 12x + 6$, $a = -1$, $b = 12$, and $c = 6$. Substituting these values into the quadratic formula, we get:

$$x = \frac{{-12 \pm \sqrt{{12^2 - 4(-1)(6)}}}}{{2(-1)}}$$

$$x = \frac{{-12 \pm \sqrt{{144 + 24}}}}{{-2}}$$

$$x = \frac{{-12 \pm \sqrt{{168}}}}{{-2}}$$

$$x = \frac{{-12 \pm 2\sqrt{{42}}}}{{-2}}$$

$$x = \frac{{12 \pm 2\sqrt{{42}}}}{{2}}$$

$$x = 6 \pm \sqrt{{42}}$$

So, the zeros of the quadratic function are $x = 6 + \sqrt{{42}}$ and $x = 6 - \sqrt{{42}}$. These are both real numbers. There are no imaginary zeros for this quadratic function.