Question

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

Answer 1

`f(x) = 6x ^(2) + 12x - 7 \\ x = ( - 6 + √(78) )/(6) \: , - ( - 6 - √(78) )/(6)`

Answer 2

Given:

The quadratic function is:

`f(x)=6x^2+12x-7`

To find:

The zeros of the quadratic function.

Solution:

Quadratic formula: If a quadratic equation is
`ax^2+bx+c=0`, then zeros of the quadratic equation are:

`x=(-b\pm √(b^2-4ac))/(2a)`

We have,

`f(x)=6x^2+12x-7`

For zeros,
`f(x)=0`.

`6x^2+12x-7=0`

Here,
`a=6,b=12,c=-7`. Using quadratic formula, we get

`x=(-12\pm √((12)^2-4(6)(-7)))/(2(6))`

`x=(-12\pm √(144+168))/(12)`

`x=(-12\pm √(312))/(12)`

`x=(-12\pm 17.6635)/(12)`

Now,

`x=(-12+17.6635)/(12)` and
`x=(-12-17.6635)/(12)`

`x=0.47195833...` and
`x=-2.47195833...`

`x\approx 0.472` and
`x\approx -2.472`

Therefore, the zeros of the given quadratic function are 0.472 and -2.472.