Mastering Systems of Equations for Success | Olympiads Prep!
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- Опубліковано 26 чер 2024
- Mastering Systems of Equations for Success | Olympiads Prep!
🏆 Ready to elevate your math game to Olympiad-level? Dive into our comprehensive guide on mastering systems of equations! 📈 In this video, we unravel the secrets behind solving complex equations with ease, equipping you with the skills needed to ace Olympiad math competitions. From fundamental strategies to advanced techniques, we cover it all to ensure your success. Whether you're a seasoned competitor or just starting your Olympiad journey, this video is your ultimate preparation tool. Get ready to conquer the Olympiad arena with confidence! 🥇 #olympiadpreparation #mathsuccess #equations #systemsofequations #mathematics #problemsolving #competitionprep #matholympiad #olympiadmathematics #mathchallenges #algebra #maths
Topics covered:
System of equations
Algebra
Logarithms
Properties of logarithm
Cubic equation
Algebraic identities
Algebraic manipulations
Vieta's Method
Solving systems of equations
Synthetic division method
Rational root theorem
Math tutorial
Math Olympiad
Math Olympiad Preparation
Timestamps:
0:00 Introduction
0:32 Properties of logarithm
2:48 Algebraic identities
5:08 Cubic equation
5:57 Synthetic division
6:46 Quadratic equation
8:50 Vieta's method
9:55 Solutions
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Thanks for watching!
@infyGyan
X= Y= 1/ 2
X=1/2 et Y=1/2.
The given can be
(1) x^3 +y^3 = 1/4
(2) xy = 1/4
By (1) and (2)
4t^3 -3t -1 =0;
(t -1)(2t +1)^2 = 0
where t = x +y.
Thus to solve
2 system of eqns
(i) x +y= 1 and xy =1/4
(ii) x +y=-1/2 and xy =1/4.
By (i),
(x, y) = (1/2, 1/2)
BY (ii)
(x, y) = cmplx sols
After some manipulations, we get x^3+y^3=1/4 and xy=1/4. we have to solve 64x^6-16x^2+1=0. Let u=x^3 and solve for u the new equation 64u^2-16u+1=0. We get twice u=1/8, then x=1/8 and y=1/8.
If I solve it in this way, does it violate any principle of mathematics?
Other View and please let me know, if anything is wrong :
Substitute p for x and q for y. Then again substitute y for p & x for q. You arrive at the same set of two equations that we begin with.
It means X and Y are both equal. So substitute Y=X in first equation. So you have X^3 = 1/8 => x = 1/2 = y.