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What are the zeros of the quadratic function f(x) = 6x2 – 24x + 1?

A) x = –2 + square root 1/2 or x = –2 – square root 1/2

B) x = 2 + square root 1/2 or x = 2 – square root 1/2

C) x = –2 + square root 23/6 or x = –2 – square root 23/6

D) x = 2 + square root 23/6 or x = 2 – square root 23/6

Respuesta :

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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.

Answer:

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

    [tex]D.\ x=2+\sqrt{\dfrac{23}{6}},\ x=2-\sqrt{\dfrac{23}{6}}[/tex]

Step-by-step explanation:

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

                [tex]f(x)=6x^2-24x+1[/tex]

We know that any quadratic function of the type:

[tex]ax^2+bx+c=0[/tex]

has solution as:

[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

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

we are asked to find x such that:

[tex]f(x)=0[/tex]

i.e.

[tex]6x^2-24x+1=0[/tex]

i.e. here [tex]a=6,\ b=-24\ and\ c=1[/tex]

Hence, we have the solution as:

[tex]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}}[/tex]

        The correct answer is: Option : D

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