Question

What are the zeros of the quadratic function f(x) = 6x^2 + 12x – 7?

Answer 1

Answer:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3} }{2}$

Step-by-step explanation:

To find the zeros of the quadratic function f(x)=6x^2 + 12x – 7 we need to factorize the polynomial.

To do so, we need to use the quadratic formula, which states that the solution to any equation of the form ax^2 + bx + c = 0 is:

$x=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

So, the first thing we're going to do is divide the whole function by 6:

6x^2 + 12x – 7 = 0 -> x^2 + 2x - 7/6

This step is optional, but it makes things quite easier.

Then we using the quadratic formula, where:

a=1, b= 2, c = -7/6.

Then:

$x=\frac{-2±\sqrt{2^{2}-4(1)(-7/6)}}{2}$

$x=\frac{-2±\sqrt{4 +14/3}}{2}$

$x=\frac{-2±\sqrt{26/3}}{2}$

So the zeros are:

$x1=\frac{-2+\sqrt{26/3}}{2}$

$x2=\frac{-2-\sqrt{26/3}}{2}$