Please help will give brainlist

2 answers:

6 0

6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. These are the zeroes.

katovenus [111]2 years ago

5 0

<h2>Answer:</h2>

The zeros of the quadratic function f(x) are:

$D.\ x=2+\sqrt{\dfrac{23}{6}},\ x=2-\sqrt{\dfrac{23}{6}}$

<h2>Step-by-step explanation:</h2>

We are given a quadratic function in terms of x as:

$f(x)=6x^2-24x+1$

We know that any quadratic function of the type:

$ax^2+bx+c=0$

has solution as:

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

Here we are asked to find the zeros of the function f(x) i.e.

we are asked to find x such that:

$f(x)=0$

i.e.

$6x^2-24x+1=0$

i.e. here $a=6,\ b=-24\ and\ c=1$

Hence, we have the solution as:

$x=\dfrac{-(-24)\pm \sqrt{(-24)^2-4\times 6\times 1}}{2\times 6}\\\\i.e.\\\\x=\dfrac{24\pm \sqrt{576-24}}{12}\\\\i.e.\\\\x=\dfrac{24\pm \sqrt{552}}{12}\\\\i.e.\\\\x=\dfrac{24\pm 2\sqrt{138}}{12}\\\\i.e.\\\\x=\dfrac{24}{12}\pm \dfrac{2\sqrt{138}}{12}\\\\i.e.\\\\x=2\pm \dfrac{\sqrt{138}}{6}\\\\i.e.\\\\x=2\pm \sqrt{\dfrac{138}{36}}\\\\i.e.\\\\x=2\pm \sqrt{\dfrac{23}{6}}\\\\i.e.\\\\x=2+\sqrt{\dfrac{23}{6}},\ x=2-\sqrt{\dfrac{23}{6}}$