A nice Math Olympiad Algebra Equation | Find the Value of x ?

Solve algebra equation #algebra#olympiad
Hello, my beloved family! 😍😍😍
I hope everyone is doing great! 😲 😲😲
If you enjoyed this video on How to solve this math problem, please show your support by liking and subscribing to my channel. Your support means the world to me! 😊😊😊
#matholympiad
#algebra​
#mathematics​
#maths
#math
#olympiad
#matholympiad #algebra#math#simplification #Exponents#vedicmath#viralmathproblem #howto#mathematics #mathematicslesson#calculus
France math Olympiad
Germany math Olympiad
Japanese math Olympiad
china math Olympiad
math Olympiad questions
maths Olympiad
France math Olympiad
math Olympiad
maths Olympiad questions
Japanese multiplication method
math
math Olympiad problems
math problem
algebra 2
nice olympiad Math Problem
Olympiad mathematics
Olympiad questions
Olympic math
Math Olympiad | A Nice Algebra Problem | How to solve this problem
A Nice Math Olympiad Algebra Problem
A Nice Exponents Problem
A Nice Math Olympiad Exponential Problem
Olympiad Algebra
Solving quartic equation
Solving A Quadratic Equations
International Math Olympiad Problem
International Math Olympiad Problem,
math olympiad topics
olympiad preparation
international olympiad maths
olympiad mathematics
international olympiads
international maths olympiad
olympiad
junior math olympiad
international mathematics olympiad
mathematics
math olympiad
international math olympiad 2024
international math olympiad
math olympiad problem
math olympiad preparation
american math olympiad question
math olympiad questions
math olympiad question
math olympiad problem
olympiad mathematics
math olympiad
math olympiad training
Canada maths olympiad
math olympiad problems
math olympiad preparation
math olympiad questions
math olympiad algebra problem
maths olympiad
olympiad, mathematical olympiad
france math olympiad question
olympiad math problems
france maths olympiad preparation
olympiad math
maths trick
math olympiad algebra
france math olympiad questions
high school math
France | Can You Solve this ? Math Olympiad
can you solve this complex problem?
can you solve this logic puzzle?
can you solve this olympiad question?
olympiad
maths olympiad
olympiad mathematics
math olympiad
math olympiad question
math olympiad questions
mathematical olympiad
math olympiad training
math olympiad preparation
math olympiad problem
olympiad math
math olympiad algebra problem
math olympiad problems
olympiad math problems
maths
france maths olympiad
Luxembourg- Math Olympiad Questions
thailand junior maths olympiad problem
olympiad math problems
beautiful algebra problem
viral math problem
math olympiad problem
Nice Algebra Math Simplification Find Value of X
Russian Math Olympiad Question.
Japanese | Can you solve this ? | Math Olympiad
Nice Exponent Math Simplification
Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this Problem?
Japanese Math Olympiad Question | A Nice Algebra Problem
UK | Can you solve this ?| Math Olympiad
Japanese Math Olympiad | Calculators Not allowed !
France Math Olympiad | Calculator Not Allowed!
Germany| Can you solve this ? |A Nice Maths Olympiad problem
China | Can you solve this ? Math Olympiad
France | Can you solve this ? | Math Olympiad
Iran Math Olympiad - Algebra - Find f(0)?
France Math Olympiad | Calculator Not Allowed
Russian- Math Olympiad Question
International Maths Olympiad Problem | Alg.Nice Algebra Math Simplification Find Value of X
Russian Math Olympiad Question.
Japanese | Can you solve this ? | Math Olympiad
Nice Exponent Math Simplification
Math Olympiad | A Nice Algebra Problem | How to solve for X and Y in this Problem?
Japanese Math Olympiad Question | A Nice Algebra Problem
France | Can you solve this ?| Math Olympiad
Japanese Math Olympiad | Calculators Not allowed !
France Math Olympiad | Calculator Not Allowed!
Germany| Can you solve this ? |A Nice Maths Olympiad problem
China | Can you solve this ? Math Olympiad
A Nice Math Olympiad Question
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
A Nice Exponents Problem
A Nice Math Olympiad Exponential Problem
viral math problem,
math Olympiad problem
Math Olympiad Question | Equation Solving| You should know this trick!!
Japanese -A Nice Math Olympiad Problem
- Math Olympiad Problem | You should know this trick!!
Viral Math Olympiad Problem | How to solve
Algebra - Nice Math Olympiad Problem #maths​
US Math Olympiad problem #maths​
Nice Exponent Math Simplification | Find the value of X??
Math Olympiad Question | Nice Algebraic Equations
math Olympiad problems ! viral math problems
Brazil Maths Olympiad Question #maths​ #superacademy247
Math Olympiad Question | Algebra
A Nice Chinese Olympiad Exponential Problem

[@vacuumcarexpo]

The fastest way of solving this type of equations often seen on TH-cam videos is adding -6x+9 to both hands.
((x-3)+(3x/(x-3)))^2=25

[@schlingel0017]

LHS using (a+b)² - 2ab: (x+3x/(x-3))² - 6*x²/(x-3) - 16 = 0. Equalize denumerator in first parantheses: (x²/x-3)² - 6*x²/(x-3) - 16 = 0 substitute x²/(x-3) = u
Now he have
u²-6u-16 = 0.
u1 = 8 => x1;2 = 4±2i√2.
u2 = -2 => x3;4 = -1± √7.

[@schlingel0017]

Write the left hand side of the first equation in form of (a+b)² - 2ab. You will be surprised how quicker it will be to solve then. I will provide detailed solution today if no one else does.

[@superacademy247]

Please do because the Math community needs more methods. Thanks 👍💯 for your support.

[@razin4419]

we can also use this method and find easily all solutions in less time..
a=x & b=3x/(x-3)

[@argenispipo63]

Mucha vuelta inútiles

[@NadiehFan]

You posted the exact same video two months ago:
th-cam.com/video/Y9YvxshF6SA/w-d-xo.html
I also posted a comment on your video at the time which presented a much more convenient and concise solution relying on the technique of converting a product of two quantities into a difference of squares and which avoids the need to deal with fractions or substitutions. For convenience sake I'll repost my comment on that video here.
Substituting
(1) y = 3x/(x − 3)
in the original equation gives
(2) x² + y² = 16
while from (1) we get
(3) 3(x + y) = xy
so the real solutions pairs (x, y) are the intersections of a circle and a hyperbola, but I don't see the advantage of an additional trigonometric substitution here. Rewriting (2) as
(4) (x + y)² − 2xy = 16
and setting the sum and product of x and y equal to s and p, that is
(5) x + y = s
(6) xy = p
and substituting (5) and (6) in (3) and (4) we have
(7) 3s = p
(8) s² − 2p = 16
From (7) and (8) we obtain
(9) s² − 6s − 16 = 0
which gives s = −2 or s = 8 and so (s, p) = (−2, −6) or (s, p) = (8, 24). Now, (s, p) = (−2, −6) means that x and y are the solutions of the quadratic equation t² + 2t − 6 = 0 which gives
x = −1 + √7 ⋁ x = −1 − √7
and (s, p) = (8, 24) means x and y are the solutions of the quadratic equation t² − 8t + 24 = 0 which gives
x = 4 + 2i√2 ⋁ x = 4 − 2i√2
A different approach which does not require a substitution and which does not require having to deal with fractions starts by multiplying both sides by (x − 3)² to get rid of the fraction. We can then proceed as follows
x²(x − 3)² + 9x² = 16(x − 3)²
x²(x² − 6x + 9) + 9x² = 16(x − 3)²
x²(x² − 6x + 18) = 16(x − 3)²
(x² − 3(x − 3) + 3(x − 3))(x² − 3(x − 3) − 3(x − 3)) = 16(x − 3)²
(x² − 3x + 9)² − 9(x − 3)² = 16(x − 3)²
(x² − 3x + 9)² − 25(x − 3)² = 0
(x² − 3x + 9 + 5(x − 3))(x² − 3x + 9 − 5(x − 3)) = 0
(x² + 2x − 6)(x² − 8x + 24) = 0
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2

[@superacademy247]

I'll create more similar questions and do presentation using your two methods. Thanks 👍💯 for your support and input

[@NadiehFan]

@@superacademy247 Yet another method is to note that (x − 3)² = x² − 6x + 9 and therefore x² = (x − 3)² + 6x − 9 so we can rewrite the equation
x² + (3x/(x − 3))² = 16
as
(x − 3)² + 6x − 9 + (3x/(x − 3))² = 16
or
(x − 3)² + 6x + (3x/(x − 3))² − 25 = 0
Here (x − 3)² + 6x + (3x/(x − 3))² is a perfect square because 6x is twice the product of (x − 3) and 3x/(x − 3) and we also have 25 = 5² so we get
((x − 3) + 3x/(x − 3))² − 5² = 0
which gives us a difference of two squares at the left hand side whereas the right hand side is zero. This allows us to factor the left hand side using the difference of two squares identity and then apply the zero product property. However, before doing so we first multiply both sides by (x − 3)² to eliminate the fraction so we get
((x − 3)² + 3x)² − (5(x − 3))² = 0
(x² − 3x + 9)² − (5x − 15)² = 0
(x² + 2x − 6)(x² − 8x + 24) = 0
x = −1 + √7 ⋁ x = −1 − √7 ⋁ x = 4 + 2i√2 ⋁ x = 4 − 2i√2

[@aninditabanerjee6862]

Simplify\(x^{2}+\left(\frac{3\cancel{x}}{\cancel{x}}-3
ight)^{2}=16\)Cancel the common factor\(x^{2}+\left(\frac{3\cancel{x}}{\cancel{x}}-3
ight)^{2}=16\)\(x^{2}+(3-3)^{2}=16\)Subtract the numbers\(x^{2}+(\textcolor{#C58AF9}{3-3})^{2}=16\)\(x^{2}+(\textcolor{#C58AF9}{0})^{2}=16\)Evaluate the exponent\(x^{2}+\textcolor{#C58AF9}{0^{2}}=16\)\(x^{2}+\textcolor{#C58AF9}{0}=16\)Add zero\(x^{2}+0=16\)\(x^{2}=16\)Move terms to the left side\(x^{2}=16\)\(x^{2}-16=0\)Replace \(16\) with \(4^{2}\)\(x^{2}-16=0\)\(x^{2}-4^{2}=0\)Use the difference of squares formula\(a^{2}-b^{2}=(a-b)(a+b)\)\(x^{2}-4^{2}=0\)\((x-4)(x+4)=0\)\((x-4)(x+4)=0\)
Solution \(x=4\\ x=-4\)