Question

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

Answer 1

we have

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

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

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

Factor the leading coefficient

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

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

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

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

Rewrite as perfect squares

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

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

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 equation are

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

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