## New Way to Solve Quadratic Equations

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Sun Dec 08, 2019 12:59 am
• I thought this was a cool find. Quadratic equations generally have at most two places where they intersect the x-axis (their roots). Math professor Po-Shen Loh figured that between the two roots there must be a number, and he solves for that number and uses it to find the roots.

https://www.technologyreview.com/s/6147 ... ions-easy/

Basically, the end result is instead of memorizing a single complicated formula, you can memorize two simpler ones (the upshot is that instead of entering two calculations into your calculator like you do with the quadratic formula, you just have to enter one).

Suppose you have the following quadratic equation:

x^2 + Bx + C = 0. (we've divided A out already)

The old way to solve this is the quadradic equation, which is

x = (-B ± sqrt[B^2 - 4C])/(2)

The way Loh does it, however, makes a little more intuitive sense, and the two formulas are a bit simpler.

As I said above, a quadratic equation has two roots. Suppose the number in between the two roots is u.

Then Loh realized that u is just

u = sqrt([B/2]^2 - C)

Very easy, just plug and chug.

After you get u, the roots are just:

x = -(B/2) ± u.

IMHO this is easier. You don't have to do two different calculations into your calculator, which for me already makes it worth the effort to learn a new method.

So in summary, here's how to use the new method to solve quadratic equations:

Step 1: Divide all the terms in the quadratic equation by the leading coefficient (A).

Step 2: Plug B and C into the following formula: u = sqrt([B/2]^2 - C)

Step 3: Plug u into the root formulas, which are -(B/2) ± u.

EDIT- I suppose an example is in order. From a random polynomial generator:

x^2+17x+72 = 0

Step 2:

u = sqrt([17/2]^2 - 72)
u = 0.5

Step 3:

x = - (17/2) ± 0.5

x = -8 and x = -9

viola.
5_Golden_Rings
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Sun Dec 08, 2019 10:06 am
• With your example equation at least, there is a quicker and simpler way: Since all the terms are positive, and since by inspection 17 is 8+9 and 72 is 8*9, then factoring the equation in your head gives the roots as (x + 8 ) and (x + 9). Since the roots are where each factor = 0, then by inspection x = -8, x = -9. If the original equation had B = 1 or B = -1 and C = -72 then the same thing would apply, except the first part would be that -72 is 8*9 with either the 8 or 9 being negative, and the roots would be at either x = -8, 9 if B = -1 or x = 8, -9 if B = 1. I haven't checked the veracity of the rest of the new method, I'll work on it later and get back to you. I might agree that this is actually pretty cool for the more complex equations, since electrical (controls) engineering uses a lot of these second-order polynomials with complex roots. Actually, so do a lot of other engineering disciplines, for instance designing suspensions for cars and deciding spring rate vs shock damping uses the same principles. You definitely want the roots to be imaginary numbers if you want your system to be inherently stable!
GeekHawk
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Sun Dec 08, 2019 5:40 pm
• GeekHawk wrote:With your example equation at least, there is a quicker and simpler way: Since all the terms are positive, and since by inspection 17 is 8+9 and 72 is 8*9, then factoring the equation in your head gives the roots as (x + 8 ) and (x + 9). Since the roots are where each factor = 0, then by inspection x = -8, x = -9. If the original equation had B = 1 or B = -1 and C = -72 then the same thing would apply, except the first part would be that -72 is 8*9 with either the 8 or 9 being negative, and the roots would be at either x = -8, 9 if B = -1 or x = 8, -9 if B = 1. I haven't checked the veracity of the rest of the new method, I'll work on it later and get back to you. I might agree that this is actually pretty cool for the more complex equations, since electrical (controls) engineering uses a lot of these second-order polynomials with complex roots. Actually, so do a lot of other engineering disciplines, for instance designing suspensions for cars and deciding spring rate vs shock damping uses the same principles. You definitely want the roots to be imaginary numbers if you want your system to be inherently stable!

Correct.....you want to imagine you are inherently stable. Seahawkfan80
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Mon Dec 09, 2019 3:55 am
• GeekHawk wrote:With your example equation at least, there is a quicker and simpler way: Since all the terms are positive, and since by inspection 17 is 8+9 and 72 is 8*9, then factoring the equation in your head gives the roots as (x + 8 ) and (x + 9). Since the roots are where each factor = 0, then by inspection x = -8, x = -9. If the original equation had B = 1 or B = -1 and C = -72 then the same thing would apply, except the first part would be that -72 is 8*9 with either the 8 or 9 being negative, and the roots would be at either x = -8, 9 if B = -1 or x = 8, -9 if B = 1. I haven't checked the veracity of the rest of the new method, I'll work on it later and get back to you. I might agree that this is actually pretty cool for the more complex equations, since electrical (controls) engineering uses a lot of these second-order polynomials with complex roots. Actually, so do a lot of other engineering disciplines, for instance designing suspensions for cars and deciding spring rate vs shock damping uses the same principles. You definitely want the roots to be imaginary numbers if you want your system to be inherently stable!

Sure, it varies from quadratic to quadratic which method will be the most efficient. I'm just saying that in comparison to the quadratic formula, this method has many advantages.

But the primary benefit with this method, as I see it, is (assuming you can't just eyeball it, like you pointed out with the specific example I chose), you only have to use the calculator once.

I'm not particularly great at arithmetic, so when I use the quadratic formula in my calculator, I usually have to enter in numbers twice to account for the positive and the negative case. But with this method, the midpoint number between the two roots has only one case, and it's the only calculation you have to do (because simply adding u to the negative of half the second coefficient requires no calculation most of the time).

So you cut the calculator grunt work in half.

As for the veracity of this method, it is definitely logically sound. You can derive it in the general case as easily as you can derive the quadratic formula in the general case.
5_Golden_Rings
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Mon Dec 09, 2019 12:40 pm
• Yeah, I ended up not checking yesterday. I had a \$h!t game to watch... GeekHawk
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Mon Dec 09, 2019 1:21 pm
• GeekHawk wrote:Yeah, I ended up not checking yesterday. I had a \$h!t game to watch... Completely understandable. But if you’re feeling lazy, the inventor of the method goes over the logic behind it in this video, showing in a particular example the motivation (it takes advantage of the difference of squares).

Of course, it looks different from what I wrote, because he’s actually working through the math while I just took what he did and put it in a formula (which is exactly the opposite of his goal, haha).

It’s not a proof, but you can see how it would be straight forward to show purely symbolically.
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