Question

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

Answer 1

Use the quadratic formula
[-12 +/- square root (12^2- (4)(6)(-7))]/2(6)
[-12 +/- square root (312)]/12

-1+(1/6) square root (78)
and 
-1-(1/6) square root (78)

Answer 2

Answer:

A quadratic equation is of the form: $ax^2+bx+c= 0$ .....[1] where

a. b and c are coefficient and x is the variable.

The solutions of the equation are;

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Given the function: $f(x) = 6x^2+12x-7$

To find the zero of this function.

Set f(x) = 0

⇒$6x^2+12x-7=0$

On comparing this with equation [1] we have;

a = 6, b = 12 and c = -7

then;

$x = \frac{-12\pm\sqrt{(12)^2-4(6)(-7)}}{2(6)}$

$x = \frac{-12\pm\sqrt{144+168}}{12}$

or

$x = \frac{-12\pm\sqrt{312}}{12}$

or

$x = \frac{-12\pm 2\sqrt{78}}{12}$

$x = \frac{-6\pm \sqrt{78}}{6}$

Therefore, the zeros of the quadratic function f(x) are;

$\frac{-6+\sqrt{78}}{6}$ and $\frac{-6-\sqrt{78}}{6}$