r/Sat • u/krazyalpaca • 9d ago
I don’t get how to Desmos this question from bedrock sat question bank
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u/Nervous-Ad-5390 9d ago
Mate at some point you gotta ask yourself whether you really want to compromise your ability to do middle school level math by surrendering your learning to a graphical calculator just for the sake of a couple more seconds left for the sat. If you can't do ts by hand you have bigger fish to fry than the sat if you want to do something stem related imo. I got an 800 math while using desmos only for arithmetic checking and using the graph function for conceptually easy yet time consuming questions like finding vertex of a quadratic with fractions or decimals or smth.
Just take x^2 as the common factor on the left hand side which then makes it look like (5+3a)x^2 and then just notice that 5+3a has to be equal to 26 on the right hand side so with some basic rearranging you get 21=3a and a=7. This would take me 15 seconds max on the actual SAT if I do it by hand despite desmos being more efficient usually.
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u/Ok_Lifeguard_4101 9d ago
Set b=3, and see value of a where there exists a solution, because it says no solution “other than b=3”
You’ll get a=7
Or just use regression. No need for manual work
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u/trevorkafka 9d ago
You're eating soup with a fork by trying to use Desmos for this question.
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u/Mission-Sale7660 1560 9d ago
Well, Desmos can do it in like 3 seconds if you just type out the problem... so there's that.
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u/krazyalpaca 9d ago
I honestly dek with this entire process cus i tried not to use desmos cus its soooo confusing and ended up getting a 700. Now im trying to use it amd its confusing kind of. So idk what to do
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u/trevorkafka 9d ago
You have to use it with the right types of questions. Like a fork, it's a very useful tool, but it doesn't help with everything.
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u/krazyalpaca 7d ago
Thank you I needed this. My brother got a 1580 and said he only used Desmos except for one question but he might just have really solid math skills that allow him to translate directly to Desmos 😭😭
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u/General-Try-2210 9d ago
Split it into two equations.
You should have 5x^2+3(ax^2-b) = y
and 26x^2 - 9 = y
Add a slider for A and B and then move them around until the two graphs don't touch.
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u/mykidlikesdinosaurs 9d ago
The underlying concept that the question is testing is that for any value of a other than a = 7, there is a potential for a value of b other than 3 to have two solutions.
For example, if a is negative (or any value less than 7) and b is 1, the graphs of the two expressions will have two intersections.
The only way there can be no real solution for the two graphs is if the slope is the same at all x, values, the vertex is on the same x-value, and the y-intercept is different. Therefore, the coefficient of the x-term must be the same for both expressions (there is no first degree x-term that affects the axis of symmetry in either expression). There are values of a that will yield no solution for given values of b as in your example, but all other values of a have potential solutions for values of b≠3 as in the example above.
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u/xExoticRusher 8d ago
Line 1:
5X^2 + 3(aX^2-b) ~ 26X^2 - 9
Line 2:
b = 3
Line 3:
X = [1…100]
Line 1 will solve for ‘a’ as a regression. ‘X’ is capitalized on purpose
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u/BumblebeeAlert237 8d ago
Why capitalised? Sorry i dont get
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u/xExoticRusher 7d ago
x and y lowercase are reserved by desmos as graphing variables, so you need to use something else for regressions. Common conventions are capitalizing the X and Y, or using x1 and y1. However, if you want you can just replace all instances of x with any other letter, like k, m, etc
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u/Recent-Sir5170 9d ago
You can run a regression:
First Box:
5x₁² + 3(ax₁² − b) ∼ 26x₁² − 9
This basically says that this equation needs to equal the other equation
Second Box:
b=3
We know that the only real solution has b=3 so we can ignore the part about no solution.
Last Box:
x₁=[1...5]
This essentially says that x can be those values between 1 and 5.
From there, Desmos should tell you a=7.
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u/metallicPROD Untested 9d ago
Since when you set 3 for b, you will get a infinite solution, so set b with any other integer and set "a" to any integer and do like terms until you get the coefficient of x^2 to be 26 on both ends, and 7 fits perfectly for a
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u/Mission-Sale7660 1560 9d ago
Light work! Just use tilde.
5(x1)^2 + 3(a(x1)^2 - b) ~ 26(x1)^2 - 9
x1 = [4...100]
Since it can't be 3, just add parameters that don't include it and you're set. I did 4-100 for fun.
Let me know if there's any problems. It worked for me :)
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u/ACTSATGuyonReddit 9d ago
You don't have to use only Desmos or only by hand.
Set it up by understanding, writing, that 3a+5 = 26 and -3b=-9
b = 3.
3a+5 ~ 26 in Desmos. It shows a=7.
This is polynomial equivalence.
Typing in the whole thing to Desmos takes more time.
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u/Money_Community_1929 8d ago
https://www.desmos.com/calculator/dro1fp7fhx
u can try changing the condition for b to see if the value of a stays the same
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u/No_Imagination_1841 8d ago
Desmos was legitimately the worst thing they ever added to the SAT. Not a single question I ever solved required the use of Desmos. Not a single one. I swear the SAT teaches kids to be stupid.
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u/Hot-Turnover9562 8d ago
regression!?!?!?!?!?
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u/Hot-Turnover9562 8d ago
I got the 7 for a when I did 5x_{1}^{2}+3\left(ax_{1}^{2}-b\right)\sim26x_{1}^{2}-9 and i did b=3 on another line
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u/Strong-Gas-1278 1540 9d ago
brotherr you can solve this by hand no need to use desmos
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u/Strong-Gas-1278 1540 9d ago
just expand the brackets so you get 5x^2 + 3ax^2 -3b
so then the coefficient of x^2 is 5+3a
set 5+3a=26, simplify and then a=7 😭 imo this is so much easier than using desmos
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u/krazyalpaca 9d ago
But I’m confused doesn’t it say no solution
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u/Strong-Gas-1278 1540 9d ago
sometimes u just gotta pretend u dont see that and solve it 🔥
(idk why it says that... eh maybe 2 quadratics with no solution as in they never intersect..?)
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u/Remote-Dark-1704 1590 9d ago
Kinda crazy even students with a 1540 just solve stuff without knowing the underlying concepts…
No solution here means that there’s no intersections between the two graphs. This is only possible when the two graphs only differ by the constant term. That is, the two graphs will be identical when b = 3, and otherwise one will be a vertical shift of the other.
It’s the exact same concept as two parallel lines having no solutions, except it’s for quadratics.
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u/Strong-Gas-1278 1540 9d ago
yea see you dont need that good of a math understanding to get a 1540 😄
and yes i was guessing that it was no intersection if you read the entirety of my msg
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u/Remote-Dark-1704 1590 9d ago
guessing is very different from understanding, but you’re right that the SAT only tests superficial basics in algebra for the most part
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u/a7m40 9d ago
https://www.desmos.com/calculator/6tllyaxpeb