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11.08.2018: One Lesson of Math - Solving Quadratic Equations by Factoring, Part 1


Today's soundtrack is Ice Dragon: Dream Dragon.


A quadratic term is a kind of polynomial; it is "written in the form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0" (p. 185).


We can solve quadratic equations by factoring by using the zero product property, which says that "[i]f the product of two numbers is 0, then either number or both numbers equal 0" (p. 185). This also applies to numbers in brackets: if the binomial factors (x + a) (x + b) = 0, then we know that either (x + a) or (x + b) gives us 0 (or perhaps both do)!

 

According to the book, to solve a quadratic equation by factoring, we first factor the trinomial, then use the zero product property. This will give us the two roots of the equation.


If we come across a quadratic equation where not all of the terms are on the same side of the equation, we need to move them all to the same side, so that 0 is on the other side.


A quadratic equation with only two terms listed on the left side of the equation (eg x² − 3x = 0) tells us right away that c = 0; thus, it is unlisted.


If we are given a radical quadratic equation, we first isolate the radical, then square both sides of the equation, then combine like terms, ensuring 0 is on the other side of the equation before solving.


Sometimes a quadratic equation will have two solutions; other times, it will have only one solution. User mvw on Stack Exchange drew up a graphic that explains the reason why nicely. Basically, it comes down to this: if the line touches the x axis once, we will get one solution. If the line crosses the x axis and then goes back again, we will get two solutions.

 

The book made mentions of "R.S." and "L.S." in their equations without explaining what they meant (I found out they were talking about the right and left sides of the equation), so I Googled it and came across this video from Khan Academy, which gave an alternative way to solve quadratic equations. In the video, they suggested plugging in the equation to the formula x = (-b ± √b² - 4ac) / 2a. The advantage of this method, they said, is that it can be used for the more complex problems that I'll come across later. I figure I might as well start learning now what I'll need later, rather than getting into habits that I'll have trouble shaking later.


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