Solving Quadratic Equations
In the blog posts a couple weeks ago, I talked about solving equations by using the "unwrapping" technique of reverse PEMDAS - get all the x's to one side, then undo everything being done to x. While that system typically works great, it doesn't work quite as intuitively in equations with multiple powers of x ("polynomial equations").
For example, suppose we were trying to use the unwrapping technique on x²+6 = -12x - 14. If we're following the same steps from the way we did linear equations, then we start by adding 12x to both sides, giving us x²+12x+6=-14. Then we need to undo everything being done to x. PEMDAS backwards tells us that we do addition and subtraction first, so we subtract 6 from both sides, but after that we run into trouble because of the multiple x's. We know multiplication should be the next step, but the "times 12" is only affecting one of the x's and not the other - if we try to undo it by dividing both sides by 12 like we normally would, then we end up with $x^2/12+x=-20/12$. But then to get rid of the "divide by 12" on x², we have to multiply by 12, which puts us right back to x²+12x=-20 again.
So now we're stuck in an infinite loop - divide by 12, multiply by 12, divide by 12, multiply by 12... We're never going to find x at this rate!
Since the unwrapping technique doesn't seem to be working due to the multiple x's, we first need to find a way to have only one x to unwrap. There are a couple different options: (1)completing the square and (2) factoring.
Option 1: completing the square
One way we can rewrite the equation with a single x is by completing the square. (If you don't remember how to do that, check out this post.) In the case of x²+12x+6=-14, we can complete the square on the left side by adding 36 and subtracting 36, giving us (x²+12x+36)+6-36=-14, which then would become (x+6)²-30=-14. Now it's in a format where we can use reverse PEMDAS to unwrap it as usual. First we add 30 to both sides, then we take the square root* of both sides, then subtract 6.
*One thing we have to watch out for, though - when we square root both sides of the equation, there are two possible answers. If ___² = 16, then the ___ could be either 4 or -4, so we need to take both of those options into account. So on the "subtract 6" step, we have both 4-6 and -4-6 as options—which means that -2 and -10 would both be answers to this question. If we only used the positive square root, then we'd only get one of the two correct answers.
Option 2: factoring
Another way we can get a single x to unwrap is by using factoring. In order to make this work, we're going take advantage of a math rule called the zero product property. This rule says that when you have several things being multiplied together to equal zero, then at least one of those things has to be zero. For example, if x·y·z=0, then that means x=0, y=0, or z=0. So, if we can find a way to rewrite our equation as (____)×(____)=0, then we can set each ___ equal to zero separately, giving us two equations we can use the unwrapping technique on.
So to do this method, we need to get all the terms to one side (so we have ____=0) , and then use factoring to rewrite that side as a product of two things. Once it's factored, we use the zero product property to set each set of parentheses equal to zero, giving us two equations that can each be solved with the unwrapping technique.
Here's how that looks on our example equation of x²+6 = -12x - 14: since we want to make one side of the equation be zero, we'll add 12x and 14 to both sides, giving us x²+ 12x +20=0. Next, we need a pair of numbers that adds to 12 and multiplies to 20, which turns out to be 10 and 2. This allows us to split the middle term and factor by grouping, turning the equation into (x+2)(x+10)=0. Now we can use the zero product property, which tells us that either x+2=0 or x+10=0. Finally, we solve each of those equations using reverse PEMDAS. For x+2=0, we subtract 2 from both sides to get x=-2, and for x+10=0 we subtract 10 on both sides to get x=-10. So, x turns out to be either -2 or -10, the same answers we got when we solved by completing the square.
So which option should I use?
When you're deciding which technique to use, the most important thing to consider is whether the problem is even factorable in the first place. The factoring technique will only work for some problems; you can't use it unless you can find a pair of numbers to split the middle term. For example, the problem 7x²-5x+4=0 can't use factoring, because there's not a pair of numbers that multiplies to 28 and adds to -5. To find the answer for that one, we would need to use completing the square.
Aside from that, deciding which method to use is a matter of personal preference—use whichever one you think will be easier for you. I personally find that factoring is usually easier, but that isn't always the case. Some problems are easier by factoring, while completing the square is easier for others. However, both methods should result in the same answer—if they don't, then it's a sign that you've done something wrong!
Here are a couple more examples. For each problem, I'll demonstrate both methods when possible.
Example: Solve the equation 4x²-40x-83=42
Method 1: Completing the square
We start by factoring out a 4 from the first two terms. Then we find h=5, so we add h², or 25, inside the parentheses and subtract 4·25 outside the parentheses. The parentheses become (x-5)², and the outside terms combine to be -183, so we have 4(x-5)² -183=42. Now we unwrap x: add 183 to both sides, divide both sides by 4, and square root both sides. (Don't forget to include both the positive and the negative square root!) Finally, we add 5 to both sides, resulting in $x= 25/2$ and $x=-5/2$ as the final answers.
Method 2: Factoring
First we subtract 42 from both sides so that we have everything on the left side and zero on the right: 4x²-40x-125=0. Next, we find a pair of numbers that adds up to -40 and multiplies to 4×-125, which is -500. The correct pair turns out to be -50 and 10, so we use that to split the middle and factor by grouping, resulting in (2x+5)(2x-25)=0. Now we can use the zero product property to get two equations: 2x+5=0 and 2x-25=0. "Unwrapping" to solve the first one gives us $x=-5/2$; repeating that with the second equation gets $x=25/2$.
Example: Solve the equation 47x²-24x-8=12x²+6x
Method 1: Completing the square
First we'll subtract 12x² and add 6x to both sides to get all the x² and x terms to the left. Next, we'll factor out 35 from the first two terms, giving us $35(x²-\frac{18}{35}x) - 8=0$. Then we find $h=\frac{-9}{35}$, so we add h², which is $\frac{81}{1225}$, inside the parentheses and subtract $35×\frac{81}{1225}$ outside the parentheses. The parentheses become $(x+\frac{9}{35})²$, and the outside terms combine to be $\frac{-361}{35}$, so we have 35(x-9/35)² - 361/35 = 0. Now we unwrap x: add 361/35 to both sides, divide both sides by 35, and square root both sides, resulting in x-9/35 = 19/35 or -19/35 Finally, we add 9/35 to both sides, giving us $x= 4/5$ and $x=-2/7$ as the final answers.
Method 2: Factoring
First we'll subtract 12x² and add 6x to both sides to get them to the left and have zero on the right. Next, we need a pair of numbers that adds up to -18 and multiplies to 35×-8, which is -280. 28 and 10 end up being the pair we need, so we use them to split the middle and factor by grouping, resulting in (7x+2)(5x-4)=0. Now we use the zero product property, giving us the equations 7x+2=0 and 5x-4=0. Solving the first one, we get $x=\frac{-2}{7}$; the second equation solves to be $x=\frac{4}{5}$.
Example: Solve the equation 6x²-5x=2
Method 1: Completing the Square
We start by factoring out 6 from the two terms on the left. Then we find $x=\frac{-5}{12}$, so we add h², or $\frac{25}{144}$, inside the parentheses and subtract $6·(\frac{25}{144})$ outside the parentheses. The parentheses become $(x-\frac{5}{12})$, and the outside term simplifies to $\frac{-25}{24}$, so we have $6(x-\frac{5}{12})² -\frac{25}{24}=2$. Now we unwrap x: add $\frac{-25}{24}$ to both sides, divide both sides by 6, and square root both sides, giving us $\frac{\sqrt{73}}{12}$ and $\frac{-\sqrt{73}}{12}$. Finally, we add $x=\frac{5}{12}$ to both sides, resulting in $\frac{5+\sqrt{73}}{12}$ and $\frac{5-\sqrt{73}}{12}$ as the final answers.
Method 2: Factoring
First we'll subtract 2 from both sides to get zero on the right side. Now we need to find a pair of numbers that adds to -5 and multiplies to 6×-2 = -12. Looking at the factor pairs of -12, none of them actually work for adding up to -5, so we can't factor this problem - we can only do it by completing the square.
Your turn! Test your knowledge by giving the following problems a try.
Use completing the square to solve for x:
13x²+78x+60=34
2x²-3x-8=2x+12
Use factoring to solve for x:
-6x²+11x+19=12
x²+2x+4=3x²+7x+1