Question

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

Answer 1

Given the equation:

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

The question is: What are the zeros of this quadratic function? this happens when f(x) = 0, so:

$f(x)=6x^{2}+12x-7=0$

a = 6
b = 12
c = -7

Then the zeros are given through equation:

$x_{12}= \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}= \frac{-12\pm \sqrt{12^{2}-4(6)(-7)}}{2(6)}$

∴ $x_{1}=0.47$
$x_{2}=-2.47$

Answer 2

Solution:

Thus, the zeros are x=0.47,-2.47

Explanation:

We have been given the function

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

In order to find the zeros of this function, we will solve the quadratic equation $f(x)=6x^2+12x-7=0$

We'll solve this by quadratic formula which is given by

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

We have

$a=6,\:b=12,\:c=-7:$

On substituting these values, we get

$x_{1,\:2}=\frac{-12\pm \sqrt{12^2-4\cdot \:6\left(-7\right)}}{2\cdot \:6}\\\\x_{1,\:2}=\frac{-12\pm \sqrt{312}}{2\cdot \:6}\\x_{1,\:2}=\frac{-12\pm 2\sqrt{78}}{12}\\\\\mathrm{The\:final\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\x=\frac{\sqrt{78}-6}{6},\:x=-\frac{6+\sqrt{78}}{6}\\\text{In decimal, we have}\\x=0.47,-2.47$

Thus, the zeros are x=0.47,-2.47