Answer-
[tex]\boxed{\boxed{\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}}}[/tex]
Solution-
The quadratic function is,
[tex]6x^2-24x + 1[/tex]
a = 6, b = -24, c = 1
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]=\dfrac{-(-24)\pm \sqrt{-24^2-4\cdot 6\cdot 1}}{2\cdot 6}[/tex]
[tex]=\dfrac{24\pm \sqrt{576-24}}{12}[/tex]
[tex]=\dfrac{24\pm \sqrt{552}}{12}[/tex]
[tex]=\dfrac{24\pm 2\sqrt{138}}{12}[/tex]
[tex]=\dfrac{12\pm \sqrt{138}}{6}[/tex]
[tex]=\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}[/tex]
The roots are of the given quadratic functions are,
[tex]x=\frac{12+\sqrt{138}}{6},\:x=\frac{12-\sqrt{138}}{6}[/tex]
We have given the quadratic function f(x) = 6x^2 – 24x + 1.
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
a=6,b=-24 and c=1 by using the given formula
[tex]x_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:6\cdot \:1}}{2\cdot \:6}[/tex]
Simplify the above equation
[tex]_{1,\:2}=\frac{-\left(-24\right)\pm \sqrt{\left(-24\right)^2-4\cdot \:6\cdot \:1}}{2\cdot \:6}[/tex]
[tex]x_{1,\:2}=\frac{-\left(-24\right)\pm \:2\sqrt{138}}{2\cdot \:6}[/tex]
[tex]x_1=\frac{-\left(-24\right)+2\sqrt{138}}{2\cdot \:6},\:x_2=\frac{-\left(-24\right)-2\sqrt{138}}{2\cdot \:6}[/tex]
[tex]\frac{-\left(-24\right)-2\sqrt{138}}{2\cdot \:6}[/tex]
[tex]=\frac{24-2\sqrt{138}}{12}[/tex]
[tex]=\frac{2\left(12-\sqrt{138}\right)}{12}[/tex]
[tex]=\frac{12-\sqrt{138}}{6}[/tex]
Therefore the roots are
[tex]x=\frac{12+\sqrt{138}}{6},\:x=\frac{12-\sqrt{138}}{6}[/tex]
To learn more about the roots visit:
https://brainly.com/question/2833285 #SPJ3
