In this section, we will review how to solve quadratic equations by different methods. This will lay a good foundation for later sections in this section.
So far, we have learned 3 methods to solve quadratic equations in the form of \(ax^2+bx+c=0\text{:}\)
When \(b=0\text{,}\) we can use the square root property. An example is \(x^2-4=0\text{.}\)
If we can easily factor the equation's left side, we should solve the equation by factoring. An example is \(x^2-4x-12=0\text{.}\)
If we cannot solve the equation by the first two methods, we can use the Quadratic Formula. Again, we can always choose to use the formula, although the first two methods could be easier.
Let's look at a few examples on how to wisely choose different methods to solve quadratic equations.
Solve for \(y\) in \(y^2-49=0\text{.}\)
Another way to write the solution is \(y=7\text{ or }y=-7\text{.}\)
We could have solved this equation by factoring:
\begin{align*} y^2-49\amp=0\\ (y+7)(y-7)\amp=0\\ y+7=0\amp\text{ or }y-7=0\\ y=-7\amp\text{ or }y=7 \end{align*}The square root method is obviously easier.
To remove the square, not only can we square root \(x^2\text{,}\) we can also square root any perfect square like \((x+1)^2\text{,}\) as we will see in the next example.
Solve for \(p\) in \(-40=10-2(p-1)^2\text{.}\)
Let's check the solution \(p=-4\text{:}\)
\begin{align*} -40\amp=10-2(p-1)^2\\ -40\amp\stackrel{?}{=}10-2(-4-1)^2\\ -40\amp\stackrel{?}{=}10-2(-5)^2\\ -40\amp\stackrel{?}{=}10-2(25)\\ -40\amp\stackrel{?}{=}10-50\\ -40\amp\stackrel{?}{=}-40 \end{align*}The solution \(p=-4\) is checked. Checking \(p=6\) is left as exercise.
When \(b\ne0\) in \(ax^2+bx+c=0\text{,}\) we cannot use the square root property any more. If we can factor the trinomial, we can solve quadratic equations by factoring. Let's look at a few examples.
Solve for \(x\) in \(x^2-4x-12=0\text{.}\)
Solve for \(x\) in \(2x^2-10x-28=0\text{.}\)
Before we factor, we must make one side of the equation \(0\text{,}\) so we can use the Zero Product Property described in Property 8.1.2. Let's look at the next example.
Solve for \(x\) in \((x+4)(x-3)=18\text{.}\)
Again, we must make the equation's right side \(0\text{,}\) so we can factor and use Zero Product Property.
\begin{align*} (x+4)(x-3)\amp=18\\ (x+4)(x-3)\subtractright{18}\amp=18\subtractright{18}\\ (x+4)(x-3)-18\amp=0\\ x^2+x-12-18\amp=0\\ x^2+x-30\amp=0\\ (x+6)(x-5)\amp=0\\ x+6=0\amp\text{ or }x-5=0\\ x=-6\amp\text{ or }x=5 \end{align*}When it's difficult or impossible to factor the trinomial in \(ax^2+bx+c=0\text{,}\) we have to resort to the Quadratic Formula:
\begin{equation*} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{equation*}We will solve Example 12.1.4 again, but with the Quadratic Formula.
Solve for \(x\) in \(x^2-4x-12=0\text{.}\)
Identify that \(a=1\text{,}\) \(b=-4\) and \(c=-12\text{.}\) We will substitute them into the Quadratic Formula:
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-12)}}{2(1)}\\ x\amp=\frac{4\pm\sqrt{16+48}}{2}\\ x\amp=\frac{4\pm\sqrt{64}}{2}\\ x\amp=\frac{4\pm8}{2}\\ x\amp=\frac{4+8}{2}\text{ or }x=\frac{4-8}{2}\\ x\amp=6\text{ or }x=-2 \end{align*}Before we use the Quadratic Formula, we must change a quadratic equation into the standard form \(ax^2+bx+c=0\) to identify the values of \(a\text{,}\) \(b\) and \(c\text{,}\) like in the next example.
Solve for \(x\) in \(4x^2+9=12x\text{.}\)
First, we convert the equation into standard form:
\begin{align*} 4x^2+9\amp=12x\\ 4x^2+9\subtractright{12x}\amp=12x\subtractright{12x}\\ 4x^2-12x+9\amp=0 \end{align*}Now we can identify that \(a=4\text{,}\) \(b=-12\) and \(c=9\text{.}\) We will substitute them into the Quadratic Formula:
\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-(-12)\pm\sqrt{(-12)^2-4(4)(9)}}{2(4)}\\ x\amp=\frac{12\pm\sqrt{144-144}}{8}\\ x\amp=\frac{12\pm\sqrt{0}}{8}\\ x\amp=\frac{12\pm0}{8}\\ x\amp=\frac{12}{8}\\ x\amp=\frac{3}{2} \end{align*}Solving Quadratic Equations with Different Methods
