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Compassionate
 one year ago
What is the quadratic formula and how do I use it? @Zale101
Compassionate
 one year ago
What is the quadratic formula and how do I use it? @Zale101

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dan815
 one year ago
Best ResponseYou've already chosen the best response.1it originates from completing the square method

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1What type of expressions would I need to use this one, is it limited to binomials?

dan815
 one year ago
Best ResponseYou've already chosen the best response.1all equations of the form ax^2+bx+ c= 0 a,b,c can be any real number including 0

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2The 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 \(\large x^2+\frac{b}{a}x+\frac{c}{a}=0\) \(\large x^2+\frac{b}{a}x+\frac{b^2}{4a}\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2\frac{b^2}{4a}+\frac{c}{a}=0\) \(\large (x+\frac{b}{2a})^2=\frac{b^2}{4a}\frac{c}{a}\) \(\large (x+\frac{b}{2a})^2=\frac{b^24ac}{4a^2}\) square root both sides \(\large x+\frac{b}{2a}=\sqrt{\frac{b^24ac}{4a^2}}\) \(\Large x=\frac{b\pm\sqrt{b^24ac}}{2a}\)

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2This is he proof for the quadratic formula. YOU don't need to do it but you can always memorize it!

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2Sorry for keeping you waiting...

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1It's okay. So what is all this talk I hear about binomials in quadratic formulas? What's the point?

Zale101
 one year ago
Best ResponseYou've already chosen the best response.2binomials in the quadratic? If you have two binomials multiplying together (x2)(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.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1Why is it if I have (x2)(x+1) the solutions really become x = 2, x = 1

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1\[s(t) = –g \times t^2 + vt + h\] i think this would be easier to remember.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2When you have \(\large\rm f(x)=(x2)(x+1)\), you're asking why this gives us "solutions" of x=2 and x=1?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2By "solutions", we mean zeroes, or where the function intercepts the xaxis. This happens when y=0.\[\large\rm 0=(x2)(x+1)\]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 (x2) or (x+1) has to be 0.\[\large\rm 0=\color{royalblue}{(x2)}\color{orangered}{(x+1)}\]So either \(\large\rm 0=x2\) (adding 2 to each side gives us x=2) or \(\large\rm 0=x+1\) (subtracting 1 from each side gives us x=1)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm 0=\color{royalblue}{(x2)}\color{orangered}{(x+1)}\]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\[\large\rm 0=\color{royalblue}{(x2)}\color{orangered}{0}\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1Scratch 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ummm, 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,\[\large\rm h=(x2)^2+3\]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,\[\large\rm h=3(x2)^2+3\]will affect the growth rate and change the shape a bit.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So for your example: \(\large\rm y=3x^2+x2\) 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

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1Sure. Pick an example. I understand that the quadratic formula finds both xintercepts, 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Just 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: \(\rm 0=(x2)(x+1)\) 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I'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..\[\large\rm f(x)=\color{orangered}{(xh)^2}+k\]When it is in this form, we call it vertex form. And the vertex of the function is located at \(\large\rm (h,k)\).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2For 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444025523865:dwI start with a square.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444025568081:dwI'm going to let this length across be x and the other part be 4.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Since it's a square, it has equal dimension on the other side,dw:1444025623649:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So now we'll fill in these smaller squares. Notice the upper left square is an \(\large\rm x~by~x\) square.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444025674721:dwAnd the other rectangular shapes have side lengths 4 and x.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So 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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2dw:1444025797639:dwOnce 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 \(\large\rm (x+4)^2\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Maybe I'm going into too much weird detail here :) lol

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm f(x)=\color{orangered}{x^2+4x+4x}+5\]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.\[\large\rm f(x)=\color{orangered}{x^2+4x+4x+16}16+5\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1I'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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Ok 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.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}16+5\]

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1Question: 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 xintercept will give me the time it touches the ground. In High School Math, are we just generally finding the time of objects?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Good good. Yes. In motion problems those are the three features you'll spend a lot of time on. Maximum height, Starting and end point.

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1In 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So in this example we're working on:\[\large\rm f(x)=\color{orangered}{\left(x+4\right)^2}11\]Our vertex is located at: \(\large\rm (4,11)\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2So that would give you your base point.dw:1444026482133:dw

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2But knowing exactly how it grows is a little tricky to graph. That's where those "solutions" come into play, ya?dw:1444026562732:dwWherever it is that they might be

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh, 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`.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm f(x)=\left(x+4\right)^211\]\[\large\rm 0=\left(x+4\right)^211\]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,\[\large\rm x=4\pm\sqrt{11}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2quadratic formula* not function.. sorry typo :(

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1What's the need to complete the square, whjy can't we just plug it into the quadratic formula, find the xintercepts, then find hte vertex?

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1I understand you're knowledgeable of the subject, but I don't know how to complete squares. You're confusing me

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Hmm. It's pretty straight forward finding the xcoordinate 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 ycoordinate 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:

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1I apprecaite your help, but it's 1:30am. Maybe we can pickup tomorrow.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Haha I was thinking the same thing XD Sorry for rambling on so long!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2I get carried away sometimes hehe

Compassionate
 one year ago
Best ResponseYou've already chosen the best response.1It's great, actually, you're enthusiastic. Never be ashamed of it. TTYL
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