Question

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

x = –1 – and x = –1 +
x = –1 – and x = –1 +
x = –1 – and x = –1 +
x = –1 – and x = –1 +

Answer 1

`x=-1+(√(78))/(6)`,
`x=-1-(√(78))/(6)`.

Explanation:

The given quadratic equation function is f(x) = 6x²+12x -7

To find the zeros of the quadratic equation we rewrite the function as

6x² + 12x - 7 = 0

Now we know the quadratic formula to get the solutions as

`x=\frac{-b\pm \sqrt{b^(2)-4ac}}{2a}`

here a = 6

b = 12

c = -7

By putting these values in the formula

`x=(-12\pm √(144-4* 6* (-7)))/(2* 6)`

`=(-12\pm √(144+168))/(12)=(-12\pm √(312))/(12)`
`=(-12\pm √(4* 78))/(12)=-1\pm (2√(78))/(12)`

`=-1\pm (√(78))/(6)`

`x=-1+(√(78))/(6)`

and
`x=-1-(√(78))/(6)`

Answer 2

we have

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

To find the zeros equate the function to zero

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

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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

Factor the leading coefficient

`7= 6(x^(2) + 2x)`

Complete the square. Remember to balance the equation by adding the same constants to each side

`7+6= 6(x^(2) + 2x+1)`

`13= 6(x^(2) + 2x+1)`

Rewrite as perfect squares

`13= 6(x+1)^(2)`

`(13/6)=(x+1)^(2)`

square root both sides

`x+1=(+/-)\sqrt{(13)/(6)}`

`x=-1(+/-)\sqrt{(13)/(6)}`

`x1=-1+\sqrt{(13)/(6)}`

`x2=-1-\sqrt{(13)/(6)}`

therefore

the answer is

The zeros of the quadratic function are

`x=-1+\sqrt{(13)/(6)}` and
`x=-1-\sqrt{(13)/(6)}`