Can you Pass Harvard University Admission Interview ?

University Admission Exam Question || Algebra Problem || Entrance Aptitude Simplification Test || Tricky Interview
Harvard University
Oxford University
Cambridge University
Ivy League Schools
#math #maths
#algebra #exponent#olympiad #simplification#exam
Oxford University Interview Question
Cambridge University Interview Question
Harvard University Interview Question
Hello, my beloved family! 😍😍😍
I hope everyone is doing great! 😲 😲😲
If you enjoyed this video on How to solve this Math Olympiad problem, please show your support by liking and subscribing to my channel.👍👍👍 Your support means the world to me! 😊😊😊
#matholympiad #algebra#math#simplification #Exponents#vedicmath#viralmathproblem #howto#mathematics #mathematicslesson
#oxford #cambridge #harvard
France math Olympiad
Germany math Olympiad
Japanese math Olympiad
china math Olympiad
The Surprising TRUTH about Harvard University Entrance Exam Nobody tells you
maths Olympiad questions
Law School Admission Test(LSAT)
math Olympiad problems
Graduate Record Exam(GRE)
math problem
algebra 2
Math Olympiad | A Nice Algebra Problem | How to solve this problem
A Nice Math Olympiad Algebra Problem
A Nice Math Olympiad Exponential Problem
Solving quartic equation
Solving A Quadratic Equations
First year Application Exam question
International Math Olympiad Problem
GPA needed to attend University
math olympiad topics
international olympiad maths
A-Leve Exam
olympiad mathematics
international olympiads
international maths olympiad
junior math olympiad
international mathematics olympiad
Computer Science mathematics
international math olympiad
math olympiad problem
american math olympiad question
olympiad mathematics
Pure Mathematics
Canada maths olympiad
Malaysia math olympiad problems
math olympiad questions
math olympiad algebra problem
University Medical Requirement Exam Interview
olympiad, mathematical olympiad
france math olympiad question
Eligibility Requirenment for medical School
france maths olympiad preparation
olympiad math
math olympiad algebra
Mortgage cost Calculation Entrance Exam in banking
france math olympiad questions
high school math
France | Can You Solve this ? Math Olympiad
Engineering Undergraduate Math Requirements
can you solve this olympiad question?
How lawyers calculate cost of litigation to get admission to the bar
maths olympiad
olympiad mathematics
math olympiad question
math olympiad training
math olympiad problem
olympiad math
math olympiad algebra problem
Loans Cost and Calculation test for agents
olympiad math problems
maths
Luxembourg- Math Olympiad Questions
olympiad math problems
beautiful algebra problem
Nice Algebra Math Simplification Find Value of X
University Software Engineering Entrance Interview
Japanese | Can you solve this ? | Math Olympiad
Nice Exponent Math Simplification
health insurance coverage exam test tricks
Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this Problem?
This is an interesting question with amazing concepts!
This is beautifully solved for you to understand #maths #viral #mathematics #exponential #olympiad
Japanese Math Olympiad Question | A Nice Algebra Problem
UK | Can you solve
Japanese Math Olympiad | Calculators Not allowed !
Life insurance policies course admission exam
France Math Olympiad | Calculator Not Allowed!
Germany| Can you solve this ? |A Nice Maths Olympiad problem
China | Can you solve this ? Math Olympiad
Undergraduate department of statistics entrance exam question
France Math Olympiad | Calculator Not Allowed
International Maths Olympiad Problem
Actuarial science entrance test
Japanese | Can you solve this ? | Math Olympiad
Nice Exponent Math Simplification
Actuarial science admission question
Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this Problem?
Japanese Math Olympiad Question | A Nice Algebra Problem
Japanese Math Olympiad | Calculators Not allowed !
Car Insurance Cost Calculation Aptitude Test
France Math Olympiad | Calculator Not Allowed!
Germany| Can you solve this ? |A Nice Maths Olympiad problem
Germany| Can you solve this ? |A Nice Math Olympiad
Math Olympiad | A Nice Algebra Problem | How to solve this problem
A Nice Math Olympiad Algebra Problem
Insurance Agents Aptitude test
A Nice Math Olympiad Exponential Problem
Math Olympiad Question | Equation Solving| You should know this trick!!
How medics calculate cost of insurance
Japanese -A Nice Math Olympiad Problem
Insurance cost calculator for undergraduates
Insurance Policies Exam for representatives | You should know this trick!!
Viral Math Olympiad Problem
Algebra - Nice Math Olympiad Problem #maths​
Actuarial Science Undegraduate Aptitude Test
Nice Exponent Math Simplification | Find the value of X??
Medical Insurance Course Admission Exam
Math Olympiad Question | Nice Algebraic Equation
Brazil Maths Olympiad Question #maths​ #superacademy247
Math Olympiad Question | Algebra
A Nice Chinese Olympiad Exponential Problem

[@kareolaussen819]

Since 17 is a prime number, this problem is particulary well suited for integer factorization into two quadratics: x^4 - 72x -17 = (x^2+px+q)(x^2-px+r)= x^4 + (q+r -p^2)x^2 -(q-r)px +qr = 0. We must have qr =-17 with q+r a positiive squared integer. Hence q=17, r=-1, p=4, which matches with (q-r)p=18*4=72. I.e (x^2+4x+17)(x^2-4x-1)=0. Hence (x+2)^2 = 4-17 =−13 => x=-2 +- sqrt(13)i or (x-2)^2 = 4+1 = 5 => x=2 +- sqrt(5).

[@TheNizzer]

Excellent method. Far, far superior to the one provided, which has no reason for embarking on masses of unnecessary algebra. Candidates at the level for such an interview are likely to be doing most of it in their head.

[@kareolaussen819]

n^3+17n=648 = 2^3*3^4 Search for positive integers n such that (n^2+15)=648/n. Possible values are n=1, 2, 3, 4, 8, 9, 18, ... n=4: 16+17?=2*81; no. n=8: 64+17?=81: yes. n=9: 81+17?=8*9=72: no. Hence factorization is (n-8)[n^2 -(a+b)n+ab] =0, where a+b=-8, ab=81. I.e n=8, or n^2 + 8n + 81 = 0

[@raghvendrasingh1289]

x^4 = 72 x + 17 (x^2+k)^2 = 2 kx^2+ 72 x + (k^2+17) We should choose k such that 2 k and k^2+17 both are perfect squares 2 k is perfect square when k = 2 , 8 , 18 , 32, 50 , 72 and so on we choose k = 8 because then k^2+17 = 81 ( perfect square ) Hence (x^2+8)^2 = (4 x+9)^2 rest follows at once

[@key_board_x]

x⁴ - 72x - 17 = 0 Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ² x⁴ - 72x - 17 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ² (x² + λ)² - 2λx² - λ² - 72x - 17 = 0 (x² + λ)² - [2λx² + λ² + 72x + 17] = 0 ← let's try to get a second member as a square (x² + λ)² - [2λx² + 72x + (λ² + 17)] = 0 → a square into […] means that Δ = 0 → let"s calculate Δ Δ = (72)² - 4.[2λ * (λ² + 17)] → then, Δ = 0 4.[2λ * (λ² + 17)] = 72² 8λ.(λ² + 17) = 72² λ.(λ² + 17) = 648 λ³ + 17λ = 648 λ³ + 17λ - 648 = 0 λ³ + 17λ - (512 + 136) = 0 λ³ + 17λ - 512 - 136 = 0 (λ³ - 512) + (17λ - 136) = 0 (λ³ - 8³) + 17.(λ - 8) = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) (λ - 8).(λ² + 8λ + 8²) + 17.(λ - 8) = 0 (λ - 8).[(λ² + 8λ + 8²) + 17] = 0 (λ - 8).(λ² + 8λ + 64 + 17) = 0 (λ - 8).(λ² + 8λ + 81) = 0 First case: (λ - 8) = 0 → λ = 8 Second case: (λ² + 8λ + 81) = 0 → we don't study these solution, because we have already: λ = 8 → it's enough Restart: (x² + λ)² - [2λx² + 72x + (λ² + 17)] = 0 → where: λ = 8 (x² + 8)² - [16x² + 72x + (64 + 17)] = 0 (x² + 8)² - (16x² + 72x + 81) = 0 ← second member as a square (x² + 8)² - (4x + 9)² = 0 → recall: a² - b² = (a + b).(a - b) [(x² + 8) + (4x + 9)].[(x² + 8) - (4x + 9)] = 0 [x² + 8 + 4x + 9].[x² + 8 - 4x - 9] = 0 (x² + 4x + 17).(x² - 4x - 1) = 0 First case: (x² + 4x + 17) = 0 x² + 4x + 17 = 0 Δ = 4² - (4 * 17) = 16 - 68 = - 52 = 52i² x = (- 4 ± i√52)/2 x = (- 4 ± 2i√13)/2 x = - 2 ± i√13 Second case: (x² - 4x - 1) = 0 x² - 4x - 1 = 0 Δ = (- 4)² - (4 * - 1) = 16 + 4 = 20 x = (4 ± √20)/2 x = (4 ± 2√5)/2 x = 2 ± √5

[@kareolaussen819]

If someone has to be repeatedly explained that a^n*b^n = (a*b)^n, or that (m+n)^2 = m^2+n^2+2m*n, I kind of doubt that such a person is interested in solving quartic equations...

[@ShriH-d1o]

as there is no X^3 term, let X^4 -72X-17= (x^2+aX+p)(X^2-aX+q); => pq= -17; p+q-a^2=0(as no X^2 term)> p+q=a^2 (1);=>p+q is +ve, now pq= -17; take p= -1 & q= 17; from(1) -1 +17= a^2; => a=±4; also -72X= (aq-ap)X; => a(q - p)=-72; as q -p= 17 - (-1)=18 ; a= - 4. Hence=> X^4 -72X -17= (X^2 -4X -1)(X^2+4X+17)=0.....

[@danielntoko2117]

Very complicated method. Use polynomial method .

[@laogui2425]

This was hideously long-winded and without explanation of why you were doing whatever it was ... If you wanted to derive two quadratic factors, it was sooooooooooo much simpler, with simple equations to derive the coefficients, made easier by 17 being a prime. You should really offer an opening explanation of what you intend to achieve: I want to find two quadratics which together give the equation. To do this i will use the method of blah blah.