What are the zeros of the quadratic function f(x) = 6x² + 12x - 7?

      x = (-9 + 2i√6)/12    and (-9 - 2i√6)/12

Step-by-step explanation:

"Finding zeroes" means find the x-values that make f(x) = 0, so we have...

0 = 6x² + 12x - 7

Use quadratic equation to solve...

x = [-12 ± √(12² - 4(6)(-7))]/[2(6)]

    x = [-12 ± √(144 + 168)]/12

        x = [-9 ± √-24]/12  

***√-24 becomes (√24)(√-1) which becomes (√4)(√6)(√-1),

√-1 = i, so this simplifies to 2i√6

            x = [-9 ± 2i√6]/12

                  x = (-9 + 2i√6)/12    and (-9 - 2i√6)/12

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

0

}

step-by-step explanation: