18.
[tex]f(x)=6x^2-3x-5[/tex]
a)
C)
[tex]\begin{gathered} f(x)=0 \\ 6x^2-3x-5=0 \end{gathered}[/tex]
Divide both sides by 6:
[tex]x^2-\frac{x}{2}-\frac{5}{6}=0[/tex]
Add 5/6 to both sides:
[tex]\begin{gathered} x^2-\frac{x}{2}=\frac{5}{6} \\ \end{gathered}[/tex]
Add 1/16 to both sides:
[tex]x^2-\frac{x}{2}+\frac{1}{16}=\frac{43}{48}[/tex]
Write the left hand side as a square:
[tex](x-\frac{1}{4})^2=\frac{43}{48}[/tex]
Take the square root of both sides:
[tex]\begin{gathered} x-\frac{1}{4}=\pm\frac{\sqrt[]{\frac{43}{3}}}{4} \\ x=\frac{1}{4}\pm\frac{\sqrt[]{\frac{43}{3}}}{4} \end{gathered}[/tex]
B)
For a equation of the form:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \end{gathered}[/tex]
We can find the roots using the following formula:
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \end{gathered}[/tex]
For:
[tex]\begin{gathered} 6x^2-3x-5 \\ a=6 \\ b=-3 \\ c=-5 \\ so\colon \\ x=\frac{3\pm\sqrt[]{-3^2-4(6)(-5)}}{2(6)} \\ x=\frac{3\pm\sqrt[]{9+120}}{12} \\ x=\frac{3\pm\sqrt[]{129}}{12} \\ x=\frac{1}{4}\pm\frac{\sqrt[]{129}}{12} \end{gathered}[/tex][tex]\begin{gathered} x=\frac{1}{4}+\frac{\sqrt[]{129}}{12}\approx1.196 \\ x=\frac{1}{4}-\frac{\sqrt[]{129}}{12}\approx-0.696 \end{gathered}[/tex]
