### Theorem 1.2.1. Quadratic Formula.

Given the second-degree polynomial equation \(ax^2 + bx + c = 0\text{,}\) where \(a\neq0\text{,}\) solutions are given by

\begin{equation*}
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\end{equation*}

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.

Given the second-degree polynomial equation \(ax^2 + bx + c = 0\text{,}\) where \(a\neq0\text{,}\) solutions are given by

\begin{equation*}
x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.
\end{equation*}

\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*}

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

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.

\(a=\) , \(b=\) , \(c=\)

\(5\)

\(-6\)

\(-8\)

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 \(a = {5}\text{,}\) \(b = {-6}\text{,}\) \(c = {-8}\text{.}\)

Using the quadratic formula, solve the equation.

\(\left\{2,\frac{-4}{5}\right\}\)

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.

This exercise tests that nested tasks work.

Consider the quadratic equation \({6x^{2}-31x-30} = 0\text{.}\)

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.

\(a=\) ,

\(6\)

Take the coefficient of \(x^2\) for the value of \(a\text{.}\) *In this case*, \(a = {6}\text{.}\)

\(b=\) ,

\(-31\)

Take the coefficient of \(x\) for the value of \(b\text{.}\) *In this case*, \(b = {-31}\text{.}\)

\(c=\)

\(-30\)

Take the constant term for the value of \(c\text{.}\) *In this case*, \(c = {-30}\text{.}\)

Using the quadratic formula, solve the equation.

\(\left\{6,\frac{-5}{6}\right\}\)

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.

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 \({2x^{2}-5x-25} = 0\text{.}\)

Identify the coefficients for the quadratic equation using the standard form from Theorem 1.2.1.

\(a=\) , \(b=\) , \(c=\)

\(2\)

\(-5\)

\(-25\)

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 \(a = {2}\text{,}\) \(b = {-5}\text{,}\) \(c = {-25}\text{.}\)

Using the quadratic formula, solve the equation.

\(\left\{5,\frac{-5}{2}\right\}\)

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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