What is the quadratic formula and how do I use it? @Zale101

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it originates from completing the square method

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What type of expressions would I need to use this one, is it limited to binomials?
all equations of the form ax^2+bx+ c= 0 a,b,c can be any real number including 0
The quadratic formula is what i believe a short cut way to solve for x. You can factor to solve for x, but when you have unfavorable quadratics, then the quadratic formula comes to handy. From where did it came from? Well, consider the standard quadratic equation: \(\large ax^2+bx+c=0\) if you tried to solve for x, you will come up with the quadratic formula. Let us try it! It \(\large ax^2+bx+c=0\) Divide everything by a \(\large x^2+\frac{b}{a}x+\frac{c}{a}=0\) \(\large x^2+\frac{b}{a}x+\frac{b^2}{4a}-\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2-\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2=\frac{b^2}{4a}-\frac{c}{a}\) \(\large (x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\) square root both sides \(\large x+\frac{b}{2a}=\sqrt{\frac{b^2-4ac}{4a^2}}\) \(\Large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
a=/= 0
This is he proof for the quadratic formula. YOU don't need to do it but you can always memorize it!
Sorry for keeping you waiting...
It's okay. So what is all this talk I hear about binomials in quadratic formulas? What's the point?
binomials in the quadratic? If you have two binomials multiplying together (x-2)(x+1), you can solve for x without needing the quadratic formula. You can simply set the two binomials to zero and solve for x. However, both methods can give you the value of x.
Why is it if I have (x-2)(x+1) the solutions really become x = 2, x = -1
\[s(t) = –g \times t^2 + vt + h\] i think this would be easier to remember.
When you have \(\large\rm f(x)=(x-2)(x+1)\), you're asking why this gives us "solutions" of x=2 and x=-1?
yes
By "solutions", we mean zeroes, or where the function intercepts the x-axis. This happens when y=0.\[\large\rm 0=(x-2)(x+1)\]From here you need to think back to basic arithmetic. If you have \(\large\rm 0=\color{royalblue}{A}\color{orangered}{B}\), then one of the numbers, A or B, has to be 0. That's the only way you can get 0 in multiplication. Applying this to our polynomial, it tells us that the factor (x-2) or (x+1) has to be 0.\[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}\]So either \(\large\rm 0=x-2\) (adding 2 to each side gives us x=2) or \(\large\rm 0=x+1\) (subtracting 1 from each side gives us x=-1)
\[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{(x+1)}\]I hope that makes some sense. We can only get zero if we have this:\[\large\rm 0=\color{royalblue}{0}\color{orangered}{(x+1)}\]or this\[\large\rm 0=\color{royalblue}{(x-2)}\color{orangered}{0}\]
I get it. That makes sense. So what about applying the gravity formula. Let me give you an example You throw a ball at the speed of 10 m/s standing on the ground. It reaches a maximum height of 50 feet. When will it reach this height? (Time) I know there's something with a vertex, or there's a vertex formula? But this seems so confusing, because sometimes I'll need to find the height, sometimes I need to find the speed -- can you shed some light on this?
Scratch that. Lets say I'm given y = 3x2 + x – 2 and I am asking to not only find the vertex, but graph it as well
Ummm, well you have a basic understanding of what a parabola looks like, yes? If I give you \(\large\rm f=(x)^2\), it's a bowl shape `opening upward`. If I give you \(\large\rm g=-(x)^2\), it's the same shape but `opening downward`. When we throw other stuff into the mix,\[\large\rm h=-(x-2)^2+3\]We have the exact same shape, we're only moving the "base of that bowl" or vertex as we call it. Having a coefficient on the squared term though,\[\large\rm h=-3(x-2)^2+3\]will affect the growth rate and change the shape a bit.
So for your example: \(\large\rm y=3x^2+x-2\) Hmm this might be a tricky one to start with if you're not familiar with `completing the square`. Maybe I can use a simpler example? :D
Sure. Pick an example. I understand that the quadratic formula finds both x-intercepts, usually two whole positive and negative numbers, and sometime it's just one positive or negative intercept, but say I wanted to find the vertex
Just a note about that factoring thing earlier, and the "solutions" that we were guaranteed. Most polynomials can not be factored nicely. Especially anything dealing with physics like your motion equation. So it's usually not very easy to turn it into something like this: \(\rm 0=(x-2)(x+1)\) That's why the Quadratic Function is amazing. It's always works. If you find that more intuitive to use rather than these weird algebraic shortcuts, then by all means. That was just a side note :) Ok back on topic. Yes I have an example.
I'm thinking of a function like this:\[\large\rm f(x)=\color{orangered}{x^2+8x}+5\]I highlighted the x's in orange because we want to `complete the square` on them. We would like to turn f(x) into something like this:\[\large\rm f(x)=\color{orangered}{(x+number)^2}+other~number\]Lemme use letters instead of words there..\[\large\rm f(x)=\color{orangered}{(x-h)^2}+k\]When it is in this form, we call it vertex form. And the vertex of the function is located at \(\large\rm (h,k)\).
For our function, Think of it like this:\[\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5\]I've split the middle term into two equal parts. I want to show you the square method. It's not very practical, but it gives a lot of intuition into this weird method.
|dw:1444025523865:dw|I start with a square.
|dw:1444025568081:dw|I'm going to let this length across be x and the other part be 4.
Since it's a square, it has equal dimension on the other side,|dw:1444025623649:dw|
So now we'll fill in these smaller squares. Notice the upper left square is an \(\large\rm x~by~x\) square.
|dw:1444025674721:dw|And the other rectangular shapes have side lengths 4 and x.
So in this example, to complete the square, fill it in the rest of the way, we need that 4 by 4 box filled. So we need 16, ya?
|dw:1444025797639:dw|Once we have that 16, this square can be simplified down into \(\large\rm (x+4)(x+4)\) those are the side lengths. Which we would further simplify to \(\large\rm (x+4)^2\)
Maybe I'm going into too much weird detail here :) lol
So umm
\[\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5\]We'll add 16 to complete our square because that's the number we needed.\[\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}+5\]But in the land of math, we're not allowed to just add 16 willy nilly like that. It's changes the function. So, in order to keep the same function, we need to balance it. A way we can do this is by subtracting 16 at the same time.\[\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}-16+5\]
I've seen thje example, and I kind of get it, but it's a little complex for me. I'm just familiar with the quadratic formula and the vertex = -b/2a thing. I really do appreciate your time and wanting to help me, it's just some things I don't get
Ok if you're more comfortable with the shortcuts, we can skip right to the meat then :) \[\large\rm f(x)=\color{orangered}{x^2+8x}+5\]\[\large\rm b=8,\qquad \frac{b}{2}=4,\qquad\left(\frac{b}{2}\right)^2=16\]So we'll add 16 to complete our square, and subtract 16 at the same time to balance it.\[\large\rm f(x)=\color{orangered}{x^2+8x+16}-16+5\]And our orange part simplifies down to: \(\large\rm \left(x+\frac{b}{2}\right)^2\) That's why I listed all three of the b parts, we'll need all of them.
\[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-16+5\]
Question: in most quadratic problem involving motion, will I generally be asked to find the time? Say, if I'm given something like 16t^2 + 96t + 200, and I solve the vertex, that will give me the time the ball reaches it's maxiumum height, and solving for the positive x-intercept will give me the time it touches the ground. In High School Math, are we just generally finding the time of objects?
Good good. Yes. In motion problems those are the three features you'll spend a lot of time on. Maximum height, Starting and end point.
In my textbooks, we never learned about completing the square. Just basic formulas and how to find them. I haven't gotten to that part in my books. I mean, I'm familiar with factoring polynomials but not completing their squares
So in this example we're working on:\[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}-11\]Our vertex is located at: \(\large\rm (-4,-11)\)
So that would give you your base point.|dw:1444026482133:dw|
But knowing exactly how it grows is a little tricky to graph. That's where those "solutions" come into play, ya?|dw:1444026562732:dw|Wherever it is that they might be
Oh, also notice... when you put it into vertex form, you'll avoid using the quadratic function altogether. You really only want to use the quadratic function from `standard form`.
\[\large\rm f(x)=\left(x+4\right)^2-11\]\[\large\rm 0=\left(x+4\right)^2-11\]It's much easier to solve for x explicitly from here. Add 11 to each side,\[\large\rm 11=(x+4)^2\]Square root,\[\large\rm \pm\sqrt{11}=x+4\]Subtract 4,\[\large\rm x=-4\pm\sqrt{11}\]
quadratic formula* not function.. sorry typo :(
What's the need to complete the square, whjy can't we just plug it into the quadratic formula, find the x-intercepts, then find hte vertex?
I understand you're knowledgeable of the subject, but I don't know how to complete squares. You're confusing me
Hmm. It's pretty straight forward finding the x-coordinate of your vertex. We were given this \(\large\rm x^2+8x\) for the x's. The vertex is located at \(\large\rm x=-\frac{b}{2}\). But finding the y-coordinate is a lot trickier. We don't have a nice shortcut like the quadratic formula. We simply have to put it into vertex form. If you search hard enough, maybe you can find a weird shortcut formula that allows you to skip `completing the square`. But I'm not familiar with any such method D:
I apprecaite your help, but it's 1:30am. Maybe we can pickup tomorrow.
Haha I was thinking the same thing XD Sorry for rambling on so long!
I get carried away sometimes hehe
It's great, actually, you're enthusiastic. Never be ashamed of it. TTYL
\c:

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