A Nice Radical Equation | Math Olympiad

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A Nice Radical Equation | Math Olympiad | Algebra

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Күн бұрын

A Nice Radical Equation | Math Olympiad | Algebra
Welcome to another exciting math challenge! In this video, we dive into a nice radical equation that's perfect for honing your algebra skills and preparing for the Math Olympiad. Watch as we break down the problem step-by-step, providing clear explanations and useful tips along the way. Whether you're a student gearing up for a competition or just love solving math puzzles, this video is for you. Let's tackle this radical equation together and sharpen our problem-solving abilities!
Time-stamps:
0:00 Introduction
1:10 Exponent laws
3:48 Solving Quartic equation
4:51 Substitution
5:45 Solving Quadratic equation using factorization
7:20 Quadratic formula
8:20 Discriminant
9:08 Solutions
9:16 Verification
We'll cover:
Key concepts and definitions
Common pitfalls and how to avoid them
Detailed example problems with solutions
Tips and tricks to solve these equations efficiently
Additional Resources:
• An Amazing Radical Cha...
• A Wonderful Factorial ...
• Overcoming Rational Eq...
• Math Olympiad Strategi...
#mathchallenge #radicalequations #matholympiad #algebra #mathskills #algebrachallenge #math #exponents
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Пікірлер: 10

@user-kp2rd5qv8g Ай бұрын

The given equation simplifies to (x^2-x)/[sqrt(x^2-x+24)] =-2. Let t = x^2-x > t/[sqrt(t+24)] = -2 > t^2-4t-96=0 > t = -8,12.If t = -8, x^2-x+8=0, which has no real roots. If t=12, x^2-x-12 =0 > x=-3,4. We can readily verify that x=-3 is a spurious solution and that x=4 is the only real solution.

@dorkmania Ай бұрын

Multiplying LHS by (x / x) and simplifying √(x² - 2x + 1)/√(x² - x + 24) = 2 / x => x√(x - 1)² = 2√(x² - x + 24) => x² - x = 2√(x² - x + 24) Substituting t = √(x² - x + 24) => t² - 24 = 2t => t² - 2t - 24 = 0 t = 6, -4 and x² - x + 24 = t² So, x² - x + 24 = 36 or x² - x + 24 = 16 x² - x -12 = 0 ( x² - x + 8 = 0 reject) x = 4 (reject x = -3)

@kassuskassus6263 Ай бұрын

After some manipulations, we have to solve the equation x⁴ - 2x³ - 3x² + 4x - 96=0, witch is (x-4)(x+3)(x² - x + 8)=0. Thus, x=4 and x=-3 as real solutions but x=-3 is rejected after checking in the original equation, and two complex ones x=(1+or-isqrt31)/2.

@tejpalsingh366 Ай бұрын

X= 4; -3 x #

@user-ny6jf9is3t Ай бұрын

Στο συνολο R χ=-3 ή χ=4. Στο συνολο των μιγαδικων επι πλεον οι ριζες χ=(1+ -ριζα31)/2

@RajeshKumar-wu7ox Ай бұрын

X=-3,4,((1+4√2 i)/2),(1-4√2 i/2))

@user-xv7xr1mu9i Ай бұрын

1/x=t

@RealQinnMalloryu4 Ай бұрын

(1)^2 ➖ (2)^2/(x)^2= {1 ➖ 4}/x^2=3/x^2 {1x+1x}/{x+x ➖} 2x^2/x^2 {3/x^2+2x^2/x^2}=6x^2/x^4 1.2x^2 1.1x2 1x^2 (x ➖ 2x+1) (1)^2=1 (1)^2/(x)^2 = 1/x^2 {1 ➖ 1/x^2}= 0+0 ➖/x^2= 1/x^2 {24x+24x ➖}/{x+ x ➖} = 48x^2/x^2 {1/x^2+48x^2/x^2}=49x^2/x^4 =12.x^2 3^4 x^2 3^2^2 x^2 3^1^1x^2 3x^2 (x ➖ 3x+2)

@herbertklumpp2969 Ай бұрын

Multiply with x ( x>0) you get Sqrt((x-1)^2) = 2*/x * sqrt( x^2-x+24) (X-1)^2 *x^2 = 4* ( x^2 - x +24) Therefore x^4 -2x^3-3x^2+4x -96=0. You find x=4 , conclude ( x-4)* ( x^3+ 2x^2+5x +24) =0 Second solution x= -3