A Nice Hexic Equation Challenge | Solving Without Expansion
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- Опубліковано 22 тра 2024
- A Nice Hexic Equation Challenge | Solving Without Expansion
Join us for a stimulating journey through hexic equations! 🧩 In this video, we take on the challenge of solving hexic equations using innovative methods that go beyond traditional expansion techniques. Discover the beauty of problem-solving without relying on expansion and witness the thrill of unraveling complex equations in a refreshing way. Whether you're a math enthusiast or simply love a good puzzle, this video is sure to captivate and inspire. Don't miss out on the fun - dive into the world of hexic equations with us! 💡 #algebra #mathchallenge #problemsolving #creativethinking #mathematics #mathskills #mathhelp #education #solvingequations #hexicequation
🔍 In this video, you'll learn:
Strategies to approach challenging Hexic equations.
Step-by-step solutions explained with clarity.
Tips and tricks to make algebraic problem-solving feel like a breeze.
Gear up for an enlightening journey into the world of algebraic challenges!
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Timestamps:
0:00 Introduction
0:54 Substitution
2:18 Sum of squares
4:48 Sum of cubes
8:28 Solving cubic equation
10:01 Synthetic division
12:24 Quadratic equation
13:20 Solutions
13:30 Verification
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@infyGyan
I couldn't see a way to do it! Thank you, this is a clever strategy.
Let a = 2x-1.The given equation becomes (a+1)^7 -(a-1)^7 = 2^7(2^7-1). Writing a^2 = u, this reduces to 7u^3 + 35u^2 + 21u -8127 = 0, whose only real solution is u = 9. Thus, a = +/-3 > x = 2,-1.
2; -1.........
X=-1 et X=2
x=-1 and x=2
X=-1 : x= 2
Hmmm... You said that you would not use the binomial expansion theorem, then proceeded to use it several times for (x+2)^2 and (x+y)^3.
So, what if we just go ahead and use the binomial expansion theorem from 0:49 ?
x^7 + (1-x)^7 = 127
x^7 + 1 -7x +21x^2 - 35x^3 +35x^4 -21x^5 +7x^6 -x^7 = 127
7x^6 - 21x^5 +35x^4 -35x^3 +21x^2 -7x = 126
x^6 - 3x^5 +5x^4 -5x^3 +3x^2 -x - 18 = 0
Now, 2 and -1 are zeroes of this expression. After synthetic division, that leaves:
y = x^4 -2x^3 +5x^2 -4x +9 = 0
Are there any additional real roots?
Note that y is a continuous function, as are its first and second derivatives. Note also that y' = 4x^3 -6x^2 -10x - 4, which is a cubic equation. It has one real root at x = 1/2. After synthetic division, that leaves y' = x^2 - x + 2 = 0, which has no additional real roots. Therefore, y has only 1 local maximum or minimum at x = 1/2.
Now, y" = 6x^2 -6x +5. At x = 1/2, y" = 3.5 and y = 129/16, which are both greater than zero. Therefore, y is upward-facing, has a global minimum at 129/16, and can never equal zero. Therefore there are no additional real roots.
So, the only solutions are x = -2 or x = 1.
Yes, it is a direct binomial expansion of (1-x)^7 which is specified there in the question.
Thanks
infyGyan ❤️
Highest power is 7, then how hexic equation ?
x^7 will cancel out from both sides, so it is hexic.