2:15 for (2), you can use this quadratic equation to solve for the 2 numbers for the rational number under the square root to split into to create a perfect square: x^2 - Ax + (B^2)/4 = 0 where A is the rational number, which is 2 and B is the irrational number, which is sqrt(3), so x^2 - 2x + (sqrt(3)^2)/4 = 0 => x^2 - 2x + 3/4 = 0 solving this gives you the number which you use to split the rational number into, in this case, x = 1.5 and x = 0.5, so (2) becomes sqrt(1.5 + 0.5 - sqrt(3))/2 sqrt((sqrt(1.5) - sqrt(0.5))^2)/2 (sqrt(1.5) - sqrt(0.5))/2 and you can continue to solve this to get the answer in the video I haven't personally done a proof for this equation since my maths teacher gave it to us to use because it works basically all the time I've used it, maybe a proof for it would be nice at 7:36, you can add both of the equation on their respective sides and solve the quadratic equation, and you will get 6 and 2 as the answer (just in case you didn't know the actual answer yet) I'm unsure if you're allowed to use the quadratic equation in your country at this stage, let alone the version above, but mine does and we use it all the time
Depends on your restrictions I suppose. √(1+√8)), for both √R≥0 and √R≤0, can't be done by staying in the reals, but can if you may traverse the complex plane.
No, because at the end when you're trying to find two numbers that have a given sum and product, most of the time the solutions are themselves irrational with square roots so it doesn't work out. But every once in a while, you can do it.
The quickest way to verify that they're all the same is just to check that they're all positive and then square them all (although you still have to rationalize the denominator in the first one). What you did was something more, showing how to turn the first one into the second one and the second one into the third one without knowing ahead of time where you were going to end up.
Is there a workbook with algebra problems like this I could work through? My mathematics is all self taught and I really want to strengthen algebra like this
Also you can do rad (2 (4 - 2 rad (3) ) , which is rad (2) * rad (4 - 2 rad (3 ), which still follows the rad (a+b - 2 rad (a*b) ) formula, the result is the same, rad(2) * (rad (3) * rad (1)) = rad(6) * rad(2)
In sec case you can just multiply top and bottom by √2 and then √(4-2√3) is equal to √(3-2√3 +1) =√((√3-1)^2) and becuase 2>√3 it is equal to √3 -1 so then multiply top and bottom by √2 and you have anser
So, the third answer is the most simplified right? It is the only one that can be approximated by hand with only simple square root of small integer approximations.
I think both (2) and (3) are considered simplified "enough" to not get you marks taken off since they resolve the radical in the denominator, but (3) is definitely "cleaner". It's also the value you'll most often see given if you look up the exact value of sin(15°). That said, if they're not the final answer we're looking for, I personally always try to avoid simplifying a term until the very end, because often you end up having to undo that simplification in later calculations.
That depends on what would one mean by "(most) simplified", which in turn depends on context. Personally, I may prefer (√3 - 1)(√2)/4 . Other equivalent expressions are √(3/8) - √(1/8) = √(0.375) - √(0.125) = (√3750 - √1250)/100 . Since it can be quickly estimated that √3750 ≈ 61 + 29/122 ≈ 61.25 √1250 ≈ 36 - 46/72 ≈ 35.34 this expression can be quickly estimated at approximately (61.25 - 35.34)/100 = (25.91)/100 ≈ 0.259
I can follow this with no issues. It's not difficult to understand. At all.......Damned if i'd remember it when sat in front of an exam paper though. 🤣
When I was at junior school & middle school math is my lovest subject (lesson) but since at high school math is disappointment subject because until now there are annoying question and formula which too hard to understand!
For you all now in the high school I just warn you to not force yourself to learn this material (like eksponen, numeric, tribal number & square root) it's too difficult & not possible to understand!
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
Evaluating sin(15 deg) with three methods: ruclips.net/video/WwecqPzgc6A/видео.html
I just fixed it! Thank you for pointing that to me!
@@bprpmathbasics welcome sir. Your videos are really very great! ❤
When he says "nested square root", I thought he was saying "nasty square root", and I was still in full agreement
This is so satisfaying. I really enjoy these kinds of maths, playing with the numbers and the function properties
My teacher does the foil method instead of using the formula a^2 - b^2, which is a little bit confusing and longer to write, thanks for the shortcut!!
2:15 for (2), you can use this quadratic equation to solve for the 2 numbers for the rational number under the square root to split into to create a perfect square:
x^2 - Ax + (B^2)/4 = 0
where A is the rational number, which is 2 and B is the irrational number, which is sqrt(3), so
x^2 - 2x + (sqrt(3)^2)/4 = 0
=> x^2 - 2x + 3/4 = 0
solving this gives you the number which you use to split the rational number into, in this case, x = 1.5 and x = 0.5, so (2) becomes
sqrt(1.5 + 0.5 - sqrt(3))/2
sqrt((sqrt(1.5) - sqrt(0.5))^2)/2
(sqrt(1.5) - sqrt(0.5))/2
and you can continue to solve this to get the answer in the video
I haven't personally done a proof for this equation since my maths teacher gave it to us to use because it works basically all the time I've used it, maybe a proof for it would be nice
at 7:36, you can add both of the equation on their respective sides and solve the quadratic equation, and you will get 6 and 2 as the answer (just in case you didn't know the actual answer yet)
I'm unsure if you're allowed to use the quadratic equation in your country at this stage, let alone the version above, but mine does and we use it all the time
Can you always unnest a (singularity) nested square root this way?
Under certain conditions. There is a Wikipedia page on nested radicals. Nested square roots have a general solution.
Depends on your restrictions I suppose. √(1+√8)), for both √R≥0 and √R≤0, can't be done by staying in the reals, but can if you may traverse the complex plane.
No, because at the end when you're trying to find two numbers that have a given sum and product, most of the time the solutions are themselves irrational with square roots so it doesn't work out. But every once in a while, you can do it.
The quickest way to verify that they're all the same is just to check that they're all positive and then square them all (although you still have to rationalize the denominator in the first one). What you did was something more, showing how to turn the first one into the second one and the second one into the third one without knowing ahead of time where you were going to end up.
Is there a workbook with algebra problems like this I could work through? My mathematics is all self taught and I really want to strengthen algebra like this
How cool is that!
Also you can do rad (2 (4 - 2 rad (3) ) , which is rad (2) * rad (4 - 2 rad (3 ), which still follows the rad (a+b - 2 rad (a*b) ) formula,
the result is the same, rad(2) * (rad (3) * rad (1)) = rad(6) * rad(2)
In sec case you can just multiply top and bottom by √2 and then √(4-2√3) is equal to √(3-2√3 +1) =√((√3-1)^2) and becuase 2>√3 it is equal to √3 -1 so then multiply top and bottom by √2 and you have anser
Why is it equal to sin 15°?
So, the third answer is the most simplified right? It is the only one that can be approximated by hand with only simple square root of small integer approximations.
I think both (2) and (3) are considered simplified "enough" to not get you marks taken off since they resolve the radical in the denominator, but (3) is definitely "cleaner". It's also the value you'll most often see given if you look up the exact value of sin(15°).
That said, if they're not the final answer we're looking for, I personally always try to avoid simplifying a term until the very end, because often you end up having to undo that simplification in later calculations.
That depends on what would one mean by "(most) simplified", which in turn depends on context.
Personally, I may prefer (√3 - 1)(√2)/4 .
Other equivalent expressions are √(3/8) - √(1/8) = √(0.375) - √(0.125) = (√3750 - √1250)/100 .
Since it can be quickly estimated that
√3750 ≈ 61 + 29/122 ≈ 61.25
√1250 ≈ 36 - 46/72 ≈ 35.34
this expression can be quickly estimated at approximately
(61.25 - 35.34)/100 = (25.91)/100 ≈ 0.259
He must edit these, as the board didn't erase itself at the very end😅
the first one was ez but the 2nd one ..... let me think for over 15 years and I wouldn't have found some of those steps
I can follow this with no issues. It's not difficult to understand. At all.......Damned if i'd remember it when sat in front of an exam paper though. 🤣
When I was at junior school & middle school math is my lovest subject (lesson) but since at high school math is disappointment subject because until now there are annoying question and formula which too hard to understand!
For you all now in the high school I just warn you to not force yourself to learn this material (like eksponen, numeric, tribal number & square root) it's too difficult & not possible to understand!
One day I had got a fever until two days because that material. Since that I decide to left this lesson!
oh yes continuity!
Extreme manipulation!
first