Stanford University Admission Interview Tricks

Stanford University Admission Interview Tricks | Find x=?

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Super Academy

Күн бұрын

Пікірлер: 23

@Cekcom Ай бұрын

11:16 Why x = (-1 + sqrt(117))/2 was rejected?

@PetrosAvesta 2 ай бұрын

Thank you for your solution, though there is a simpler way and that's: X= (30 + (30 +X)^1/2)1/2 or X= (30 + (30 + (30 +.....)^1/2)^1/2)^1/2 Square both sides of the equation and you get: X^2 = 30 + X Or X^2 - X -30 = 0 This equation has two roots 6 & -5. -5 is not acceptable as X must be > 0. so X=6.

@davidseed2939 Ай бұрын

I did by guessing. X^2-30= sqrt(x+30) Must be positive Smallest integer x is 6 to keep left hand side positive and evaluate to 6, which is also rhs so that is the solution. Interesting that X^2-30=x is also the solution to x=sqrt(x+30). Does this generalise so that for any problem where “30” is replaced by another integer say “c” Then the solution to that problem would be the solution to the quadratic x^2-x-c=0 Ie x=(1+-sqrt(1+4c))/2 Other “easy” values of c found from c=x^2-x for integer values of x x=2,3,4,5,6,7, c = 2,6,12,20,30,42. ( note that this sequence is the triangular numbers times 2)

@PetrosAvesta Ай бұрын

@@davidseed2939 The answer to your question is "Yes". As long as C>0, the Delta for the quadratic equation X^2 - X -C = 0 is greater than zero and the equation has two real solutions. The sequence of Cs (2 times the triangular numbers) is undoubtedly generated from the integer values of X(excluding 1) sequence in the eq. C=X^2-X.

@sergeilyubski852 29 күн бұрын

Sorry could you please explain. If I square both sides I get 30+ (30+x)^1/2 = x^2 . Right ? So according to you 30 + (30 +x)^1/2 = 30 +x i.e. (30+ x)^1/2 = x . How ? Could you please explain your simplification?

@PetrosAvesta 27 күн бұрын

@@sergeilyubski852 Hi, Let's start with this equation: X = (30 + (30+X)^1/2)^1/2 Now replace the X on the RHS of the equation with the original X and repeat this for a number of times. You get the following series equation, however please note the last term in all such equations always remain to be X. Therefore we obtain: X = (30 + (30 + (30 +.......+ (30+X)^1/2)^1/2 Now square both sides and you get: X^2 = 30 + (30 + ............+(30+X)^1/2)^1/2 OR : X^2 = 30 + X.

@Penndennis Ай бұрын

A quadratic in terms of 30 - constant and variable at the same time - I love it. Thank you so much!

@superacademy247 Ай бұрын

You're very welcome! Wonderful! Glad you liked it ✅👌🙏🙏🤩💕

@jendamatus Ай бұрын

Except that I didn't guess, the solution is easy when you deal with numbers

@prollysine 2 ай бұрын

we get , x^4-60x^2-x+870=0 , (x-6)(x^3+6x^2-24x-145)=0 , x=6 , x^3+6x^2-24x-145=0 , (x+5)(x^2+x-29)=0 , x= -5 , / x^2+x-29=0 , roots not integer , not a solu , / , test , x=6 , V(30+V(30+6))=6 , V(30+6)=6 , 6=6 , OK , x=-5 , V(30-5)=5 , V(30+5)=V35 --> not 5 , x= -5 , not a solu , solu , x=6 ,

@serhiizalutskyi5911 2 ай бұрын

too complicated. video is low quality. paper is shaking.

@9허공 2 ай бұрын

let y = √(30 + x) => squaring given equation 30 + y = x^2 and 30 + x = y^2 subtracting, y - x = x^2 - y^2 => x^2 - y^2 + (x - y) = (x - y)(x + y +1) = 0 => (case y = x) x^2 = 30 + y => x^2 - y - 30 = (x - 6)(x + 5) = 0 => x = 6 (since x > 0) (case y = -1 - x) x^2 = 30 -1 - y => x^2 + x + 29 = 0 => no real solution.

@marcofrigerio2217 2 ай бұрын

m+x=y^2, m+y=x^2, x^2-y^2+x-y=0, (x-y)(x+y+1)=0. But x>0 and y>0, so x+y+1>0. From x-y=0, y=x, m+x=x^2, x^2-x-30=0 (m=30): x=6 and x=-5 (not valid).

@johnlv12 2 ай бұрын

genius

@superacademy247 2 ай бұрын

Thanks. I'm humbled 🙏🙏🙏

@whoff59 2 ай бұрын

Just tried and only some seconds of thinking needed: 6 is at a solution. Perhaps there are more solutions because of roots and resulting quadratic equations ...

@jendamatus 2 ай бұрын

6 too much easy

@CrYou575 Ай бұрын

Well intelligent guessing usually is for that particular root.