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*}
Section 1.2 The 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*}
Checkpoint 1.2.2. 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.
Checkpoint 1.2.3. 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.
Checkpoint 1.2.4. 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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