I did it a little bit differently. After simplifying the inner bracket, I did the following: (1-h)^2 + (h/2 - 1/2)^2 = (1-h)^2 + 1/4*(h -1)^2 = 125/4 Then I use the fact that (a-b)^2 = (b-a)^2 to simplify it to (h-1)^2 + 1/4*(h-1)^2 = 5/4*(h-1)^2 =125/4 (h-1)^2 = 25 Because of the square, we now have two equations to work with: h - 1 = 5 and h - 1 = -5 Solving for both, we get the same answer of h = 6 and h = -4
This is a nice example of how a quite complicated looking expression can be simplified fairly easily. This often happens in Maths, particularly in things like exams when answers are generally quite simple (to facilitate quicker and more accurate marking). Personally I'd multiply by 4 and distribute (as 2) inside the brackets before expanding but I realise this could confuse some. Later on I feel it would be simpler to just divide by 5 rather than factoring it out but that's a minor difference.
Not to be argumentative. Engineering students like me have taken Calculus and the equivalent of a second Calculus coursework but need by my Junior University math a third Calculus set of coursework to solve engineering problems. Watching this video shows types of additional Calculus brainwork engineering students go through in our problems. He definitely is talking about basic easy Calculus problems in this one to spur by Senior University math coursework thoughts involved to pass Senior year math courses to graduate properly from the University whether or not we seek Masters and Doctorate mathematics we propose to University teacher colleagues.
I'd love to see a series in which he solves final exams from highschool from all around the world! Like this comment if you would like to see this series happen! I can provide the final exam from Romania if interested. Love your content❤
@Meowcat-kr6tc It is useful in certain cases to factor out rather than just dividing through, in order to preserve information that would otherwise be lost. However, as we're only looking for the roots, we could divide through by 5 and simplify the expression. It's also worth mentioning that we are able to divide through by 5 because it is never 0. We can then simplify into two cases, because when one root is equal to 0, the other is strictly non-zero and you can effectively divide through by that factor as well.
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I did it a little bit differently. After simplifying the inner bracket, I did the following:
(1-h)^2 + (h/2 - 1/2)^2 = (1-h)^2 + 1/4*(h -1)^2 = 125/4
Then I use the fact that (a-b)^2 = (b-a)^2 to simplify it to
(h-1)^2 + 1/4*(h-1)^2 = 5/4*(h-1)^2 =125/4
(h-1)^2 = 25
Because of the square, we now have two equations to work with:
h - 1 = 5 and h - 1 = -5
Solving for both, we get the same answer of
h = 6 and h = -4
Your approach seems much simpler to me.
I noticed this different approach like you
I think you can factor out -1/2 from (1/2h-1/2)^2 at 1:25 so it becomes 1/4(1-h)^2. I think you can get a shorter answer that way😊
I did all the math by hand, which included a lot of simplifying, and used the AC method to get (5)(h - 6)(h + 4) = 0, from which I got h = {6, -4}.
I just subscribed to this channel because I love this guy's stuff
This is a nice example of how a quite complicated looking expression can be simplified fairly easily. This often happens in Maths, particularly in things like exams when answers are generally quite simple (to facilitate quicker and more accurate marking).
Personally I'd multiply by 4 and distribute (as 2) inside the brackets before expanding but I realise this could confuse some. Later on I feel it would be simpler to just divide by 5 rather than factoring it out but that's a minor difference.
Just do some factorisation at 1:20 to skip all that expansion
Wait a minute.... (½h-½)² is actualy (1-h)²/4
Why didn’t you just factor out -(½)?
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How is this calculus 3 😅
Not to be argumentative. Engineering students like me have taken Calculus and the equivalent of a second Calculus coursework but need by my Junior University math a third Calculus set of coursework to solve engineering problems. Watching this video shows types of additional Calculus brainwork engineering students go through in our problems. He definitely is talking about basic easy Calculus problems in this one to spur by Senior University math coursework thoughts involved to pass Senior year math courses to graduate properly from the University whether or not we seek Masters and Doctorate mathematics we propose to University teacher colleagues.
It’s not the focus of Calc 3 but likely something that popped up in a Calc 3 exam
I'd love to see a series in which he solves final exams from highschool from all around the world!
Like this comment if you would like to see this series happen!
I can provide the final exam from Romania if interested.
Love your content❤
u forget number 5
No
5 will never be zero, so either h-6 is zero or h+4 is zero, or both. That's why he doesn't consider a case where 5=0
@@sylqsnoop4827 why did he not just divide both sides by 5, surely the factorising is redundant when the equation is equal to 0
@Meowcat-kr6tc It is useful in certain cases to factor out rather than just dividing through, in order to preserve information that would otherwise be lost. However, as we're only looking for the roots, we could divide through by 5 and simplify the expression.
It's also worth mentioning that we are able to divide through by 5 because it is never 0. We can then simplify into two cases, because when one root is equal to 0, the other is strictly non-zero and you can effectively divide through by that factor as well.
@@Meowcat-kr6tc yes. That's what I would have done.