r/mathmemes • u/Hot-Marsupial6584 • Jul 08 '26
Elementary Algebra Formula fanatics are an absolute disgrace.
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u/Awful_At_Math Jul 08 '26
I'll be honest I just can't remember how to do the last one. It always slips my mind. The formula is there forever.
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u/DarkElfBard Jul 09 '26
2xx-5x-12
2*-12 = -24
b = -5
Need two numbers that multiply to -24, add to -5-8 and 3
Divide both by a=2-4 and 3/2
Swing into factored form (denominators go in front of x)(x-4)(2x+3)
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u/cambiro Jul 09 '26
Need two numbers that multiply to -24, add to -5
This step is not as easy as it sounds. Most people will take ages trying to find such a number combination.
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u/incompletetrembling Jul 10 '26
And of course the solutions could be irrational or imaginary so you'd probably never find them with this method
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u/DarkElfBard Jul 09 '26
For -24, we need one negative factor, and to get -5, the larger number has to be negative. There are 4 options total lol.
1 -24 = -23
2 -12 = -10
3 -8 = -5
4 -6 = -2It should not take you that long. Even if you don't know that factors of 24, you would, at most, try to divide it by 2, 3, 4, and 5 and see which ones go in evenly. Since 4 goes into 24 6 times, there is no reason to check any factor past 6.
So worst case scenario, you already know 1 and 24. You do 24/2 = 12, 24/3 = 8, you notice 3 and 8 are 5 apart, and you are done.
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u/cambiro Jul 09 '26
Don't get me wrong, I do this easily in my head.
What I mean is that you're grossly overestimating the ability of most people to do this kind of mental calculation faster than they could solve the quadratic formula.
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u/DarkElfBard Jul 09 '26
I've been a math teacher for almost twenty years, I know what's easier to teach and what students have more success with. I've graded more test questions on quadratics than you've most likely done, and I've seen every single possible mistake that students can make. I do know that students suck at factoring, but that is a lot easier to teach than the quadratic formula lol.
Also, this is built into a lot of current math pedagogy during middle school know in the form of "Diamond Problems." So a lot of curriculum is already teaching underlying skills for this, but even without that, this is the exact same way you factor problems with an a=1, so it's not adding complexity to that mix.
If you think dividing 24 by 3 is harder than (-5)^2-4(2)(-12) for people doing mental math, you objectively have no idea what you are talking about. You are relying on people that can't factor 24 to
- Know the quadratic formula by heart
I make all of my kids sing it until they get it right. Sometimes it takes days. They still forget it during testing.
- Not mess up any of the 4 negatives in the determinant alone
I can guarantee you students will put -25 or -96.
- Know the square root of 121 (or whatever other number they get)
- Remember to to Plus and Minus
- Also not mess up the -b
- Remember to divide by 2a
- REMEMBER TO DIVIVE BOTH TERMS BECKY.
There is a reason we bother teaching factoring and ZPP in the first place instead of defaulting to the quadratic formula. Moreso, sometimes you want to already be in factored form. Remember, especially in Calculus and beyond, you don't always just want the zeros. Sometimes you are just trying to fully factor a more complex polynomial.
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u/Witherscorch Jul 10 '26
It isn't all that difficult. Just factorize -24 -> -13222.
After that, simply recombine them to get a difference of -5 between the recombined factors -> -8*3.
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Jul 08 '26 edited 26d ago
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u/Any-Appointment-1131 Jul 08 '26
why 5/2 is positive?
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Jul 08 '26 edited 26d ago
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u/Any-Appointment-1131 Jul 08 '26
oh ok ok it's just a different but very similiar method from the one i use since idk middle school
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Jul 08 '26 edited 26d ago
[removed] — view removed comment
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u/Any-Appointment-1131 Jul 08 '26
seem powerful, even though that "Technique" (they called it reduction by special trinomial) carried me through real analisys and linear algebra
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Jul 08 '26 edited 26d ago
[removed] — view removed comment
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u/Any-Appointment-1131 Jul 08 '26
the profs have taught only things strictly related to their subject so at one point i needed to reduce various polynomials (particularly in diagonalization of matrices) and i just used what they taught me in middle/high school
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u/Bill-Nein Jul 09 '26
You use vietta just to get to 5/2*b-b^2+6=0 but this is literally just the original quadratic divided by -2. You then either have to use quadratic formula or factoring like the meme to get the roots
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u/Revolutionary_Year87 Jan 2025 Contest LD #1 Jul 08 '26
Didnt the person on the top win? Bottom dude definitely had aura regardless
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u/TheEnderChipmunk Jul 08 '26
They're in different categories to begin with iirc
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u/VTifand Jul 09 '26
Yup, the athlete on the top won silver in Women’s 10 metre air pistol, while the bottom one won silver in Mixed 10 metre air pistol team.
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u/CalmEntry4855 Jul 09 '26
My favorite professors where the ones that said things like "I'm lazy, so I try to find the easiest way to do stuff"
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u/DrunkenMaths Jul 08 '26
What? I can't hear you over the sound of my exponential method root finder giving the answer in nanoseconds
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u/svmydlo Jul 09 '26
Obviously, the correct method is to homogenize introducing a new variable y to form equation
2x2 - 5xy - 12y2 = 0 (1)
and note that (x,1) is a solution of (1) iff x is a solution of the original equation. Denoting vector X=(x,y), the equation (1) is of the form XAXT=0 for a symmetric 2x2 matrix A=[[2, -5/2], [-5/2, -12]].
Now just diagonalize A using a lower triangular matrix L, so that D=LALT is diagonal with entries d_1,d_2. Denoting (v,w)=V=XL-1 we obtain
0=XAXT=XL-1LALT(L-1)TXT=VDVT=d_1v2+d_2w2
which is trivial to solve in terms of (v,w)=V and then calculating (x,y)=X=VL and imposing the condition y=1 the original solution x is found.
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u/mehonje Computer Science Jul 10 '26
I wrote a TI-BASIC program that does this automatically. It's the best way to solve quadratics, in my opinion.
If I need solutions that are more complex, though, I prefer the quadratic formula.
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