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

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Answer 1 · 2018-01-13T12:30:14

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]

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