Question

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

Answer 1

Answer:

$x=-1+\frac{\sqrt{78} }{6}$ and

$x=-1-\frac{\sqrt{78} }{6}$

Step-by-step explanation:

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

To find out the zeros of the quadratic function, we apply quadratic formula

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

From the given f(x), the value of a=6, b=12, c=-7

Plug in all the values in the formula

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

$x=\frac{-12+-\sqrt{312}}{2(6)}$

$x=\frac{-12+-2\sqrt{78}}{2(6)}$

Now divide each term by 12

$x=-1+-\frac{\sqrt{78} }{6}$

We will get two values for x

$x=-1+\frac{\sqrt{78} }{6}$ and

$x=-1-\frac{\sqrt{78} }{6}$

Answer 2

It an expression or a way of saying f(x)=6x2+12-7