Take the coefficient of \(x^2\) for the value of \(a\text{,}\) the coefficient of \(x\) for \(b\text{,}\) and the constant for \(c\text{.}\)In this case, they are \(a = {5}\text{,}\)\(b = {-6}\text{,}\)\(c = {-8}\text{.}\)
1.2.2.bUse the Quadratic Formula.
Answer.
\(\left\{2,\frac{-4}{5}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {5}\text{,}\)\(b = {-6}\text{,}\) and \(c = {-8}\text{,}\) so the results are:
\begin{equation*}
x = {\frac{-\left(-6\right)+\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {2}
\end{equation*}
or
\begin{equation*}
x = {\frac{-\left(-6\right)-\sqrt{\left(-6\right)^{2}-4\cdot 5\cdot \left(-8\right)}}{2\cdot 5}} = {-{\frac{4}{5}}}
\end{equation*}
Checkpoint1.2.3.Nested tasks.
1.2.3.aIdentify Coefficients.1.2.3.a.i
Answer.
\(6\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{.}\)In this case, \(a = {6}\text{.}\)
1.2.3.a.ii
Answer.
\(-31\)
Solution.
Take the coefficient of \(x\) for the value of \(b\text{.}\)In this case, \(b = {-31}\text{.}\)
1.2.3.a.iii
Answer.
\(-30\)
Solution.
Take the constant term for the value of \(c\text{.}\)In this case, \(c = {-30}\text{.}\)
1.2.3.bUse the Quadratic Formula.
Answer.
\(\left\{6,\frac{-5}{6}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {6}\text{,}\)\(b = {-31}\text{,}\) and \(c = {-30}\text{,}\) so the results are:
\begin{equation*}
x = {\frac{-\left(-31\right)+\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {6}
\end{equation*}
or
\begin{equation*}
x = {\frac{-\left(-31\right)-\sqrt{\left(-31\right)^{2}-4\cdot 6\cdot \left(-30\right)}}{2\cdot 6}} = {-{\frac{5}{6}}}
\end{equation*}
Checkpoint1.2.4.Copy a Problem with Tasks.
1.2.4.aIdentify Coefficients.
Answer1.
\(2\)
Answer2.
\(-5\)
Answer3.
\(-25\)
Solution.
Take the coefficient of \(x^2\) for the value of \(a\text{,}\) the coefficient of \(x\) for \(b\text{,}\) and the constant for \(c\text{.}\)In this case, they are \(a = {2}\text{,}\)\(b = {-5}\text{,}\)\(c = {-25}\text{.}\)
1.2.4.bUse the Quadratic Formula.
Answer.
\(\left\{5,\frac{-5}{2}\right\}\)
Solution.
Recall that the quadratic formula is given in Theorem 1.2.1.
You already identified \(a = {2}\text{,}\)\(b = {-5}\text{,}\) and \(c = {-25}\text{,}\) so the results are:
\begin{equation*}
x = {\frac{-\left(-5\right)+\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {5}
\end{equation*}
or
\begin{equation*}
x = {\frac{-\left(-5\right)-\sqrt{\left(-5\right)^{2}-4\cdot 2\cdot \left(-25\right)}}{2\cdot 2}} = {-{\frac{5}{2}}}
\end{equation*}
1.3Open Problem Library
Checkpoint1.3.1.Cylinder Volume.
Answer1.
\(360\pi\ {\rm m^{3}}\)
Answer2.
\(1130.97\ {\rm m^{3}}\)
Solution.
We use \(r\) to represent the base’s radius, and \(h\) to represent the cylinder’s height.
A cylinder’s volume formula is \(V= (\text{base area}) \cdot \text{height}\text{.}\) A cylinder’s base is a circle, with its area formula \(A = \pi r^{2}\text{.}\)
Putting together these two formulas, we have a cylinder’s volume formula:
\(\displaystyle{ V= \pi r^{2} h }\)
Throughout these computations, all quantities have units attached, and we only show them in the final step.
If \(\sqrt{2}\) were rational, then \(\sqrt{2}=\frac{p}{q}\text{,}\) with \(p\) and \(q\) coprime. But then \(2q^2=p^2\text{.}\) By the Fundamental Theorem of Arithmetic 1 , the power of \(2\) dividing the left side is odd, while the power of \(2\) dividing the right side is even. This is a contradiction, so \(\sqrt{2}\) is not rational.
Checkpoint1.6.2.Choose one.
Answer.
\(\text{The Fundamental ... of Calculus}\)
Solution.
The correct answer is The Fundamental ... of Calculus.
Checkpoint1.6.3.Choose a Subset of Options.
Answer.
\(\text{Choice 2, Choice 4, Choice 5}\)
Solution.
The correct answer is Choice 2, Choice 4, Choice 5.
Checkpoint1.6.4.Choose a Subset of Options with Automated Labeling.
Answer.
\(\text{B, C, D}\)
Solution.
The correct answer is B, C, D.
Checkpoint1.6.5.Choose a Subset of Options with Explicit Labeling.
Answer.
\(\text{TACO, SUSHI, PIZZA}\)
Solution.
The correct answer is TACO, SUSHI, PIZZA.
1.7Tables
Checkpoint1.7.1.Complete this Table.
Answer1.
\(24\)
Answer2.
\(24\)
Answer3.
\(48\)
Answer4.
\(48\)
Solution.
\(\times\)
\(6\)
\(6\)
\(4\)
\(24\)
\(24\)
\(8\)
\(48\)
\(48\)
1.8Graphics in Exercises
Checkpoint1.8.1.A static <latex-image> graph.
Answer.
\(\mathop{\rm C}\nolimits\!\left(n+1,2\right)\hbox{ or }\frac{\left(n+1\right)n}{2}\)
Checkpoint1.8.2.A randomized <latex-image> graph.
Answer.
\(48\ {\rm cm^{2}}\)
Checkpoint1.8.3.A <latex-image> graph affected by <latex-image-preamble>.
Answer.
\(-3, 0, 3\)
Checkpoint1.8.5.Solve using a graph.
Answer.
\(\left\{1\right\}\)
Solution.
The graph reveals that the solution set to \(f(x)=1\) is \({\left\{1\right\}}\text{.}\)
Exercises
1.8.1.
Answer.
\(5\)
Solution.
1.8.2.
Answer.
\(13\)
Solution.
2Technical Examples 2.1PGML Formatting and Verbatim Calisthenics