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# Integrating WeBWorK into Textbooks: Sample Exercises

## Section1.2The Quadratic Formula

In the previous section, we saw relatively simple WeBWorK questions. This section demonstrates how even very complicated WeBWorK problems can still behave well.
Here is a theorem that gives us a formula for the solutions of a second-degree polynomial equation. Note later how the WeBWorK problem references the theorem by its number. This seemingly minor detail demonstrates the degree to which WeBWorK and PreTeXt have been integrated.

### Proof.

\begin{align*} ax^2 + bx + c &= 0\\ ax^2 + bx &= -c\\ 4ax^2 + 4bx &= -4c\\ 4ax^2 + 4bx + b^2 &= b^2 - 4ac\\ (2ax + b)^2 &= b^2 - 4ac\\ 2ax + b &=\pm\sqrt{b^2 - 4ac}\\ 2ax &=-b\pm\sqrt{b^2 - 4ac}\\ x &=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \end{align*}

### Checkpoint1.2.2.Solving Quadratic Equations.

Consider the quadratic equation $${5x^{2}-6x-8} = 0\text{.}$$

#### (a)Identify Coefficients.

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.
$$a=$$ , $$b=$$ , $$c=$$
Answer 1.
$$5$$
Answer 2.
$$-6$$
Answer 3.
$$-8$$
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 $$a = {5}\text{,}$$ $$b = {-6}\text{,}$$ $$c = {-8}\text{.}$$

#### (b)Use the Quadratic Formula.

Using the quadratic formula, solve the equation.
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*}
This conclusion is just here for testing.

### Checkpoint1.2.3.Nested tasks.

This exercise tests that nested tasks work.
Consider the quadratic equation $${6x^{2}-31x-30} = 0\text{.}$$

#### (a)Identify Coefficients.

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.
##### (i)
$$a=$$ ,
Answer.
$$6$$
Solution.
Take the coefficient of $$x^2$$ for the value of $$a\text{.}$$ In this case, $$a = {6}\text{.}$$
##### (ii)
$$b=$$ ,
Answer.
$$-31$$
Solution.
Take the coefficient of $$x$$ for the value of $$b\text{.}$$ In this case, $$b = {-31}\text{.}$$
##### (iii)
$$c=$$
Answer.
$$-30$$
Solution.
Take the constant term for the value of $$c\text{.}$$ In this case, $$c = {-30}\text{.}$$

#### (b)Use the Quadratic Formula.

Using the quadratic formula, solve the equation.
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*}
This conclusion is just here for testing.

### Checkpoint1.2.4.Copy a Problem with Tasks.

We are testing copying the quadratic equation problem above (Checkpoint 1.2.2), since it is structured with <task>, and we also provide a new seed.
Consider the quadratic equation $${2x^{2}-5x-25} = 0\text{.}$$

#### (a)Identify Coefficients.

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.
$$a=$$ , $$b=$$ , $$c=$$
Answer 1.
$$2$$
Answer 2.
$$-5$$
Answer 3.
$$-25$$
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 $$a = {2}\text{,}$$ $$b = {-5}\text{,}$$ $$c = {-25}\text{.}$$

#### (b)Use the Quadratic Formula.

Using the quadratic formula, solve the equation.
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*}
This conclusion is just here for testing.
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