Solving that Berkeley mini math tournament radical equation
ฝัง
- เผยแพร่เมื่อ 20 ส.ค. 2024
- Solving that Berkeley mini math tournament radical equation.
The original part: • A triple square root p...
Get a 2.718 shirt: blackpenredpen...
Use "WELCOME10" for 10% off
Get an Euler's Identity t-shirt 👉 bit.ly/eulersidentity
Use "WELCOME10" for 10% off
-----------------------------
For more official SAT practice problems, check out 👇
• SAT math questions fro...
Download & try the official SAT practice tests on College Board here: 👇satsuite.colle...
-----------------------------
If you find my channel helpful and would like to support it 💪, then you can
support me on Patreon: 👉 / blackpenredpen
or buy a math shirt or a hoodie: 👉 bit.ly/bprp_merch
-----------------------------
"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Feel free to leave your questions in the comment!
Subscribe for future algebra tutorials 👉 bit.ly/just_al...
-----------------------------
#justalgebra
In the beginning you said that, because of squaring everything, you needed to check each solution. X=0, x=20 are obvious. But the other two not so much. They aren’t leading to negative roots, but I feel you should probably have checked it ;).
I find it easier to square both sides and then postulate u=20x-x^2.
Then you square again and get 4u=u^2
Pretty self-forward from that point on.
Aww... Someone's in love. It must be a real crush able to knock out math knowledge out of your head
At the end you designated two solutions as X₄ :D
Why didn’t you check if all proposed solutions fall within the domain?
They do all fall within domain
@@legittaco4440 They DO, but he did not CHECK
AAAAAAAAAAAwesome!!!!!!!!!!!!!!
Cool video as always BPRP #Yay!
Probably it should be checked that 4√6 doesn't exceed the bounds 0 and 20 (it doesn't, 4√6 is 9.8)
Well we saw it as √96 first so we already knew it was
x=0 or 20, it's easy to see~
CosX
Is there any solution for |sin(x)| = |sin(180x/π)
@Afnaan2701 I thought not trivial one
If i didn't made any errors, then the following should be true:
|sin(x)| = |sin(180x/π)|
sin(x) = +/-sin(180/PI x)
sin(x) = sin(+/-180/PI x)
sin(x + 2 PI n) = sin(+/-180/PI x); n is an integer
x + 2 PI n = +/-180/PI x
2 PI n = +/-180/PI x - x
2 PI n = (+/-180 - PI)/PI x
2 PI^2 n / (+/-180 - PI) = x
White Chalk, Pink Chalk?
I didnt understand how you did the factoring at 3:30
It's just y^2-4y=0 to y(y-4)=0, but y is x(20-x)
Epic
Why do certain solutions have to have certain numbers for x?
look at the original part in the description
first comment 😁
Second comment
@@luizao9005 first comment to first comment to first comment. Only have to use the idea of unit and successor.
I've called the channel 8 news. They'll be here to interview you soon.
@@bobh6728 😁
@@chitlitlah They ain't here yet 🙂