Consider the following quadratic equation: \(x^2-5x-10=0\). The only option to solve this equation, is to try and factor. However, there are no factors of \(-10\) that will add up to \(-5\). The method of square roots cannot be applied here either, the left side of the equation is in standard form, not vertex-form. How then, can this equation be solved? A new method called the Quadratic Formula. This is an amazing formula developed that only requires the constants from the standard form of \(a,\;b\) and \(c\): $$x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$ The following outlines general steps to solve using this formula, applying them to the above example:
| General Steps | Applied to \(x^2-5x-10=0\) |
|---|---|
| Identify \(a,\;b\) and \(c\): (Watch positives & negatives!) | $$a=1,\;b=-5,\;c=-10$$ |
| Carefully plug values into formula: $$x=\frac{-\color{green}{b}}{2\color{blue}{a}}\pm\frac{\sqrt{\color{green}{b}^2-4\color{blue}{a}\color{red}{c}}}{2(\color{blue}{a})}$$ |
$$x=\frac{-(\color{green}{-5})}{2(\color{blue}{1})}\pm\frac{\sqrt{(\color{green}{-5})^2-4(\color{blue}{1})(\color{red}{-10})}}{2(\color{blue}{1})}$$ |
| Simplify under the square root: | $$\begin{array}{l} \sqrt{(-5)^2-4(1)(-10)} \\ \implies \sqrt{25-(-40)} \\ \implies \sqrt{25+40} \\ \implies \sqrt{65} \end{array}$$ |
| Simplify the fractions (if needed): | $$x=\frac{5}{2}\pm\frac{\sqrt{65}}{2}\;\color{green}{\checkmark}$$ |
Not so bad, right? One major problem involves the calculation that happens under the square root: \(b^2-4ac\), this number is called the discriminant. In the example above, the discriminant is \(65\). The discriminant has impacts on the solutions to the equation. The following outlines the possibilities:
Consider the equation: \(4x^2-8x=5\). Recall, the equation MUST equal zero: \(4x^2-8x-5=0\). Note that \(a=4,\;b=-8\) and \(c=-5\). The discriminant looks like: \((-8)^2-4(4)(-5)=64-(-80)\rightarrow144\), which is positive AND a perfect square. The finished solution would then look like: $$\begin{array}{l} x=-\frac{-8}{2(4)}\pm\frac{\sqrt{144}}{2(4)} \\ \implies x=\frac{8}{8}\pm\frac{12}{8} \\ \implies x=1\pm\frac{3}{2} \\ \implies x=1+\frac{3}{2}\;\& \; x=1-\frac{3}{2} \\ \implies x=\frac{5}{2}\;\color{green}{\checkmark}\;\& \;x=-\frac{1}{2}\;\color{green}{\checkmark} \end{array}$$
Below are 4 more examples with a video that goes through each problem. Try each example first, then watch the video to check solutions.