ANSWER:
B.
[tex]x=\frac{5+\sqrt{17}}{4}[/tex]
C.
[tex]x=\frac{5-\sqrt{17}}{4}[/tex]
EXPLANATION:
Given:
[tex]2x^2+1=5x[/tex]
To find:
The two values of x that are roots of the equation
Let's subtract 5x from both sides of the equation;
[tex]\begin{gathered} 2x^2+1-5x=5x-5x \\ 2x^2-5x+1=0 \end{gathered}[/tex]
Recall that a quadratic equation is generally given in the below form;
[tex]ax^2+bx+c=0[/tex]
Comparing both equations, we can see that;
[tex]\begin{gathered} a=2 \\ b=-5 \\ c=1 \end{gathered}[/tex]
Let's go ahead and use the below quadratic formula to solve for the values of x;
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex][tex]\begin{gathered} x=\frac{-(-5)\pm\sqrt{(-5)^2-4*2*1}}{2*2} \\ \\ x=\frac{5\pm\sqrt{25-8}}{4} \\ \\ x=\frac{5\pm\sqrt{17}}{4} \\ \\ x=\frac{5+\sqrt{17}}{4},\frac{5-\sqrt{17}}{4} \end{gathered}[/tex]
