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Quadratic equations are equations with the form
$$a{x}^{2}+bx+c=0.$$ |
The expression $a{x}^{2}+bx+c$ is called a quadratic expression, because the highest power of any of the terms is 2. There are four methods for solving quadratic equations by hand:
The results of these methods will always be the roots of the function.
Formula
The quadratic formula can be used with all quadratic expressions. The roots are
$$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}.$$ |
Quadratic equations can have no solutions, one solution or two solutions.
${b}^{2}-4ac<0\Rightarrow $ no real solutions,
${b}^{2}-4ac=0\Rightarrow $ one real solution,
${b}^{2}-4ac>0\Rightarrow $ two real solutions.
Example 1
Solve the equation ${x}^{2}+11x=-30$.
First, you have to put all the nonzero terms on one side of the equal sign to get $0$ alone on the other side:
$$\begin{array}{llll}\hfill {x}^{2}+11x& =-30\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}+11x+30& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Next, you use the quadratic formula with $a=1$, $b=11$ and $c=30$:
$$\begin{array}{llll}\hfill x& =\frac{-11\pm \sqrt{\phantom{\rule{-0.17em}{0ex}}{\left(11\right)}^{2}-4\cdot 1\cdot 30}}{2\cdot 1}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-11\pm \sqrt{121-120}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-11\pm \sqrt{1}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-11\pm 1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Set up the expressions with both the positive square root and the negative square root:
$$\begin{array}{llllllll}\hfill {x}_{1}& =\frac{-11+1}{2}\phantom{\rule{2em}{0ex}}& \hfill {x}_{2}& =\frac{-11-1}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{-10}{2}\phantom{\rule{2em}{0ex}}& \hfill & =\frac{-12}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =-5\phantom{\rule{2em}{0ex}}& \hfill & =-6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$That means the solutions are ${x}_{1}=-5$ and ${x}_{2}=-6$.
Rule
When $c=0$, the expression looks like this:
$$a{x}^{2}+bx=0$$ |
Example 2
Solve the equation $2{x}^{2}+4x=0$.
You factorize by taking $2x$ out of the parentheses:
$$2x\phantom{\rule{-0.17em}{0ex}}\left(x+2\right)=0.$$ |
Then you set up an equation for each factor:
$$\begin{array}{llllllll}\hfill 2x& =0\phantom{\rule{2em}{0ex}}& \hfill x+2& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$That will give you ${x}_{1}=0$ and ${x}_{2}=-2$.
Rule
When $b=0$, the expression looks like this:
$$a{x}^{2}+c=0$$ |
Example 3
$$\begin{array}{llll}\hfill 3{x}^{2}-27& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 3{x}^{2}& =27\phantom{\rule{1em}{0ex}}|\xf73\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}& =9\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =\pm 3\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$
Rule
You have $a{x}^{2}+bx+c=0$. When you solve by inspection, you follow these two rules:
$$\begin{array}{llll}\hfill c& =p\cdot q,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill b& =p+q.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$Here, ${x}_{1}=-p$ and ${x}_{2}=-q$ are the roots of the function, making them the solutions to the equation.
Example 4
Solve the equation ${x}^{2}-x-56=0$.
You can see that $b=-1$ and $c=-56$. You need to find values for $p$ and $q$ that can get you the solution to the equation. There are multiple combinations of factors that give you the product $-56$. Here are some of them:
$$\begin{array}{llll}\hfill -\phantom{\rule{-0.17em}{0ex}}\left(2\cdot 28\right)& =-56,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -\phantom{\rule{-0.17em}{0ex}}\left(4\cdot 14\right)& =-56,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill -\phantom{\rule{-0.17em}{0ex}}\left(7\cdot 8\right)& =-56,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$As all of these products are $-56$, they are all candidates for the solution. For each of these products, we want a negative difference between the factors, because you are looking for the answer $-1$, a negative number. That means we set the differences up like this:
$$\begin{array}{llll}\hfill 2-28& =-26\ne -1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 4-14& =-10\ne -1\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 7-8& =-1,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$That shows you that $p=7$ and $q=-8$ fits with the equation. According to the formula, that means
$${x}_{1}=-7\text{and}{x}_{2}=-\phantom{\rule{-0.17em}{0ex}}\left(-8\right)=8,$$ |
because you have to change the sign to find the solution.