Respuesta :
Given the quadratic equation:
[tex]2x^2-6x+1[/tex]The x-intercepts are the values of (x) which make the equation equal to zero
Or, the x-intercepts are the values of intersection with the x-axis
So, we will equate the equation to zero then solve it for (x):
[tex]\begin{gathered} 2x^2-6x+1=0 \\ \end{gathered}[/tex]As we will not be able to factor the equation, we will use the general formula:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]From the equation:
[tex]a=2,b=-6,c=1[/tex]substitute with the values of {a, b, c} into the general formula:
[tex]\begin{gathered} x=\frac{6\pm\sqrt[]{(-6)^2-4\cdot2\cdot1}}{2\cdot2}=\frac{6\pm\sqrt[]{28}}{4} \\ \\ x=\frac{6+\sqrt[]{28}}{4}=\frac{6+2\sqrt[]{7}}{4}=\frac{3}{2}+\frac{1}{2}\sqrt[]{7} \\ or \\ x=\frac{6-\sqrt[]{28}}{4}=\frac{6-2\sqrt[]{7}}{4}=\frac{3}{2}-\frac{1}{2}\sqrt[]{7} \end{gathered}[/tex]So, the x-intercepts are:
[tex]x=\mleft\lbrace\frac{3}{2}-\frac{1}{2}\sqrt[]{7},\frac{3}{2}+\frac{1}{2}\sqrt[]{7}\mright\rbrace[/tex]