Question

How do I do B? Solve for f(x)=0? I think I can complete the square but that is the question in part C. Is there another way to solve f(x)=0 besides completing the square?

Answer 1

18.

`f(x)=6x^2-3x-5`

a)

C)

`\begin{gathered} f(x)=0 \\ 6x^2-3x-5=0 \end{gathered}`

Divide both sides by 6:

`x^2-(x)/(2)-(5)/(6)=0`

Add 5/6 to both sides:

`\begin{gathered} x^2-(x)/(2)=(5)/(6) \\ \end{gathered}`

Add 1/16 to both sides:

`x^2-(x)/(2)+(1)/(16)=(43)/(48)`

Write the left hand side as a square:

`(x-(1)/(4))^2=(43)/(48)`

Take the square root of both sides:

`\begin{gathered} x-(1)/(4)=\pm\frac{\sqrt[]{(43)/(3)}}{4} \\ x=(1)/(4)\pm\frac{\sqrt[]{(43)/(3)}}{4} \end{gathered}`

B)

For a equation of the form:

`\begin{gathered} ax^2+bx+c=0 \\ \end{gathered}`

We can find the roots using the following formula:

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \end{gathered}`

For:

`\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=(1)/(4)\pm\frac{\sqrt[]{129}}{12} \end{gathered}`
`\begin{gathered} x=(1)/(4)+\frac{\sqrt[]{129}}{12}\approx1.196 \\ x=(1)/(4)-\frac{\sqrt[]{129}}{12}\approx-0.696 \end{gathered}`