The simpler quadratic formula | Ep. 1 Lockdown live math

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Recommends user774777 · 2020-04 · Mathematics Stack Exchange · on “What is the role played by the difference of the two roots in solving a quadratic equation?”

You might find the video at the bottom of my post useful; he explains in good detail how the difference in two roots can be found by using the following equation: $$m\pm\sqrt{m^2-p}$$ $m$ : The midpoint between the roots. $p$ : The constant which has been adjusted. $\sqrt{m^2-p}$ is the distance from the midpoint to the nearest root; obviously there will be a $\pm$ before it as since there are two roots, one will occur in the negative $x$ direction and one in the positive $x$ . The video below will explain in more detail; go to the 16th minute if you need an answer quick. Quadratic Equation Video [this video]

Recommends shivam jalotra · 2020-06 · Mathematics Stack Exchange · on “How does factoring a quadratic equation relate to the parabola on the graph and to common sense?”

Suppose a function $F: R -> R$ has n-roots ${r1, r2, r3 ... rn}$ then the function F can be written as $F(X) = (X - r1)*(X - r2)*(X - r3) ... *(X - rn)$ . So basically when we factor a function $F(x)$ in simpler degree-1 terms like $(X - ri)$ we are actually finding root ri that contributes to the function F. But why is thing true ? Lets see what remainder theorem is : Suppose : $F(x) = G(x) + rem$ Now dividing by $(X - r1)$ on both sides gives: $F(x)/(x - r1) = Q(x) + rem/(x - r1)$ $=> F(x) = Q(x)(X - r1) + rem $ If $X = r1$ , then $F(r1) = rem$ . So we can see if $rem = 0$ or $F(r1) = 0$ , then $r1$ is a root of F(x). Similarly $Q(x)$ can be factored sown to get all the remaining roots. Re

Recommends Mark Helms · 2020-07 · Mathematics Stack Exchange · on “Need help with Alternative Factoring method”

I was working on some factoring, as I have always been terrible at it, when I found 3B1B's video [this video] on an easier method. There's a TL:DR at the bottom if you're familiar. The basics are as follows: Imagine the graph of a quadratic. $x^2 - 1$ for example. It's got 2 roots $r$ and $s$ the same distance $d$ apart from a midpoint $m$ . The method only works for equations that look like $x^2 +b-c=0$ , so if you've got an $a$ , scale everything else down by dividing everything by $a$ . So now you've got that: $r+s=b$ and $r \cdot s = c$ . We also know that: $r=m-d$ and $s=m+d$ . We can do some algebra to the above to realize the following: $m=\frac{-b}{2}$ and $d = \sqrt{m^2-c}$ So neat! we'v

Recommends Doug Fir · 2020-10 · Mathematics Stack Exchange · on “Find the zeros of $f(x)=x^3−4x^2+x−4$”

I am to find the zeros and multiplicities of $f(x)=x^3−4x^2+x−4$ . The solution provided in the answers section of my book is 4 with multiplicity 1. I arrived at $2\pm\sqrt(8)$ . My working: $$x^3-4x^2+x-4$$ $$x(x^2-4x+1)-4$$ Then, focusing on the quadratic in the middle, I used 3BlueOneBrownVideo [this video] (start from ~23 minutes) to find the zeros: $$m = \frac{-b}{2}=\frac{4}{2}=2$$ $$d^2=m^2-p=2^2+4=8$$ $$r,s=m\pm\sqrt{d^2}$$ $$r,s=2\pm\sqrt{8}$$ The zeros I arrive at are therefore $2\pm\sqrt{8}$ How can I arrive at 4 per the solution? Granular baby steps much appreciated.

Recommends Joe · 2021-05 · Mathematics Stack Exchange · on “Significant Proofs Understandable to Middle School Students with Pre-Algebra Background”

Here a couple of examples I can think of: The sum of the counting numbers from $1$ to $100$ is $5050$. [LINK] The ambitious student might be able to find the general formula for the sum of an arithmetic series, provided they are given some guidance. If $a$ and $b$ are two irrational numbers, then it is possible for $a^b$ to be rational. [LINK] If $n^2$ is even then $n$ is even. This is a nice introduction to proof by contraposition. You could also use contraposition to argue that at least one digit in the decimal representation of $\pi$ occurs infinitely often. [LINK] * There are infinitely many prime numbers [LINK] .** Geometric demonstration of completing the square to solve $x^2+10x=39$ .

Cites Doug Fir · 2020-08 · Mathematics Stack Exchange · on “Zeros of polynomial functions: Find the dimensions of the box with width w, Length 2w and height w+2”

I'm reading through a textbook chapter section on zeros of polynomial functions. The exercise I am on: Find the dimensions of the box described (inches): The length is twice the width and the height is the width + 2. The volume is 192. The solution is provided as $8*4*6$ inches. I am unable to arrive at this. My working: $l=2w$ , $w=w$ , $h=w+2$ $$v=lwh$$ $$192=2ww(w+2)$$ $$192=2w^3+2w^2$$ $$0=2w^3+2w^2-192$$ $$0=w^3+w^2-96$$ I tried two approached here. First I used the rational zero theorem to come up with candidate zeros. Since my leading coefficient is 1 after my last step of dividing everything by 2, my potential zeros are factors of -96: $\pm 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96$ U

Cites Doug Fir · 2020-09 · Mathematics Stack Exchange · on “Find the inverse of $x^2+4x, x\ge-2$ using 3b1b quadratic shortcut”

I am to find the inverse of $x^2+4x$ where $x\ge-2$ . Rewriting x,y: $$x=y^2+4y$$ This is a quadratic. Watching 3BlueOneBrown's video starting at ~ 23 minutes and 30 seconds [this video] , he shows a shortcut for calculating the quadratic.Using this method I get: $$m = \frac{-b}{2}=\frac{-4}{2}=-2$$ $$d^2=m^2-p=-2^2-0=4$$ $$r,s=m\pm\sqrt{d^2}=-2\pm2=-4,0$$ Since I have the condition that $x\ge-2$ I know I must take only the positive sqrt so 0. The solution in my textbook to the inverse of $x^2+4x$ where $x\ge-2$ is $\sqrt{x+4}-2$ How can I arrive at this solution? I am unable to get there with my working?

Cites Watson · 2024-01 · Mathematics Stack Exchange · on “Why we need to know how to solve a quadratic?”

As explained in this video ( [this video] ) around 3.30, the quadratic formula is heavily used in computer graphics (method of "ray tracing").