a nested square root problem that gave calculus students trouble
ฝัง
- เผยแพร่เมื่อ 20 ส.ค. 2024
- There are tons of algebra in calculus and it is usually the case that many calculus students make mistakes in the algebra. Here's an example where we have to use the conjugate twice in order to simplify the nested square root expression.
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We wanted to get rid of nested roots and ended up with another nested root
Nice
I just love the “cancel cancel” line, feels like you’re killing unnecessary numbers off in a good way.
I love this kind of solutions and problems. Kinda challenged my algebra power. (Anyway, I didn't skip ads for you. You know, just a little appreciation for a nice work🙂🙂.)
I wonder if students often forget to list the conditions for the domain of x? You didn't list the condition in the final result since it wasn't the focus, but it does exist.
Haven't dealt too much with nested square roots, but I'm going to assume the domain would be all real x at least -9.
@@johnathanpatrick6118 The domain question actually isn't related to the square roots. When the (x - 7) on the numerator and denominator are cancelled out, that produces a removable discontinuity where x =/= 7 according to the original expression.
If you want non-complex outputs, then you are right in the conditions including x >= -9, but the whole condition for the domain would be "x >= -9, x =/= 7".
@@PennyLapin How stupid of me to forget that x-7 in the denominator... 🤦🏾♂🤦🏾♂
Thank you sir
It was an intelligent solution👌🌸
Two things:
B) and C) can be quickly ruled out because they don't have the correct powers of x on the top and bottom of the fraction, so you can do the multiple choice at a glance. You're gonna be looking for something which has a total of x^-3/4 in the fraction. A) has that, but B) is x^-1/4 and C) is x^1/4.
Can you really call the solution a "simplification"? It still has nested radicals in it.
First you have state the domain of x. x cannot be igual to -7. A simpler way to simplify this is to break it up into fractions. The expression becomes sqrt( 5 +sqrt(×+9))/(×+7) -3/(×+7)
Then square the first term you get
(5+sqrt(x+9))/(×+7)^2 -3/(×+7)
Using the same procedure the expression becomes.
Sqrt(×+9)(x +7)^2 -3/(x+7) +5/(×+7)^2
Same procedure result in
(X+9)/(×+7)^4 -3 /(×+7) -5/(×+7)^2
Which is the same as
1/(×+7)^3 +2/(×+7)^4 -3/(×+7)- 5/(×+7)^2
So we got rid of all the radicals..
How is that simplified?
I actually did this one
So, you got rid of the sqrt from the numerator to generate it in the denominator. :-)
Algebra doesn’t really depend on calculus, it’s more the opposite. So calculus students may not be better than algebra students at algebra.
The answer is A, was pretty easy honestly
Ya pero que me sirve esa hvda
It doesn’t look simplified to me. It looks like a mess in the denominator!
What I have to do for catcher yogur attention?
Maaan.. again the same mistake. You you answer do have a value when x=7, but the original fraction doesnt 🙄🙄🙄
Youre doing rationalization in reverse, whats the point?
lo
@@English_shahriar1 who asked
u r rationalizing the numerator. why u do not rationalize denominator?
Because the denominator is fine
Cause if you check the answers, none is rationalized. So it's better to leave it like that until you get so far, like in the video
First :)
Just let x=0 then only one will be right. Can't be bothered for all this algebra