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Section 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.

\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 given by

\begin{equation*} {4x^{2}-21x-18} = 0\text{.} \end{equation*}

First, identify the coefficients for the quadratic equation using the standard form from Theorem 2.1.

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

Answer 1

\(4\)

Answer 2

\(-21\)

Answer 3

\(-18\)

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 \(a = {4}\text{,}\) \(b = {-21}\text{,}\) \(c = {-18}\text{.}\)

Using the quadratic formula, solve \({4x^{2}-21x-18}=0\text{.}\)

\(x=\) or \(x=\)

Answer 1

\(6\)

Answer 2

\(-{\textstyle\frac{3}{4}}\)

Solution

Recall that the quadratic formula is given in Theorem 2.1.

You already identified \(a = {4}\text{,}\) \(b = {-21}\text{,}\) and \(c = {-18}\text{,}\) so the results are:

\begin{equation*} x = {{-\left(-21\right)+\sqrt{\left(-21\right)^{2}-4\cdot 4\cdot \left(-18\right)}} \over 2({4})} = {6} \end{equation*}

or

\begin{equation*} x = {{-\left(-21\right)-\sqrt{\left(-21\right)^{2}-4\cdot 4\cdot \left(-18\right)}} \over 2({4})} = {-{\textstyle\frac{3}{4}}} \end{equation*}