While waiting for the exam to start, the class is currently reviewing the subject on the board. Not to even mention how odd (in a good way) it is to be in a class where everyone seems to want to succeed, I'm realizing how easy it is for me to get confused between approaching a quadratic function and a quadratic equation. Funny thing; after the exam, most of my classmates seemed pretty happy with how they did. Except we all forgot that when solving a radical equation, we have to put our answers back into the original equation to check they both work. And, I officially suck at drawing parabolas. Mine ended up looking as thick as the u-bend beneath the sink. :( So, fingers crossed for a decent grade at least.
Anyway! The differences between a quadratic function:
=ax^2+bx+c $)
and a quadratic equation:
They look the same, right? I mean, one just says  $)
instead of 
, right? Hold those horses!
Sure, they both create the same result. When x is 0, f(x) will also be 0. But, the differences arise when trying to complete the square for either, because the actual purposes are different. With functions, we're trying to complete the square in order to be able to graph it. We can take our resulting equation in vertex form, figure out our vertex, y intercepts, and draw the graph. With a quadratic equation, we're trying to solve for x. Funnily enough, we sometimes do both when trying to graph a function to find our x-intercepts (or, in other words, what our x values are when y equals 0).
The key is remembering purpose. If given a quadratic expression, and told to solve for x, set it equal to zero and go through the completing the square normally. Or, cheat and use the quadratic formula. If given the same expression and told to graph it, set it equal to zero, replace the zero with f(x), and solve it with the vertex method of completing the square. It sounds more complicated than it is.
Below for math students and math geniuses who wish to tell me I made a mistake (please do), my personal guide to completing the square. FYI, these will always be put at the bottom of my posts, so that people not interested in it don't have to wade through it all to get to the interesting stuff about how I failed my exam and decided to quit maths in pursuit of a career with the French foreign legion.
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Quadratic Function |
Quadratic Equation |
Start here!
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Isolate or move the
constant (c) term: |
=(ax^2+bx) \qquad \quad +c $)
(By associative property)
|
(Subtract c from both sides)
|
Remove the leading x^2
coefficient - it's easier to
complete the square without it! |
=a(x^2+\frac{b}{a}x) \qquad \quad +c $)
(Factor a out) |
(Divide both sides by a) |
Look at what was our b term.
Take it, half it,
|
\cdot(\frac{1}{2})=\frac{b}{2a} $) |
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Take that, and square it.
This is, for want of a better
phrase, the magic number
|
^2=\frac{b^2}{4a^2} $) |
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Add it to make our perfect
square quadratic
|
=a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}) \qquad \quad +c-(a)(\frac{b^2}{4a^2}) $)
(Now this is where it's tricky. You have to add that magic number to the inside the parentheses. But, the rule of equality does not apply here. You can't add it to the other side as well, to keep the equation balanced. So, we need to use the additive identity of 0.
What that means, is that we can add zero to one side of the equation without changing its value. Sowhatever we add inside the parentheses, we have to subtract the same value outside of them.
The trick is that whatever you add inside, you have to remember that you're really adding that magic number, with the factored out a. So that's why when we subtract it on the outside, we have to multiply it by a before we subtract it in order to keep the function perfectly balanced. Go ahead and multiply it out to test it. ) |
(Here, just add the magic number to both sides of the equation. Remember the LCD to combine these two constant terms)
(\frac{4a}{4a})+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$)
(So you end up with):
|
Take the quadratic and rewrite
it as a quantity squared.
|
=a(x+\frac{b}{2a})^2 \qquad \quad +k $)
(k is the simplification of the two constant terms that were outside the formula.) |
^2=\frac{b^2-4ac}{4a^2}$) |
Square root both sides of the
equation.
|
But not on the function side! We've taken a standard form quadratic function, and turned it into vertex form. That's as far as we need to go, here!
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The denominator simplifies once it's been square rooted. |
Isolate the x term.
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And look! It's the quadratic formula! |
It looks so very complicated. It's actually more difficult when you just look at the algebra form. If you're reading this because you're trying to learn how to complete the square, it's better to practice with actual numbers. So, here's a reasonably simple one. For the function, take it from standard form and put it into the vertex form. For the equation, solve for x. Let me know how you get on!
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