Crack the Challenge: Mastering An Incredible Algebra Equation

Crack the Challenge: Mastering An Incredible Algebra Equation

Embark on a mathematical journey as we delve into the world of algebraic challenges! In this video, we're tackling an incredible equation that will test your problem-solving skills. Join us to unlock the secrets, master the steps, and crack the challenge together. Get ready to elevate your algebraic prowess!

Topics Covered:
1. Understanding the basics of challenging radical equation from algebra.
2. Analyzing the quartic equation using algebraic tricks.
3. Step-by-step approach to solving the algebraic quartic equation.
4. Tips and tricks for handling tricky radicals like a pro.
5. Algebraic identities and manipulations while solving equations.

Timestamps:
0:00 Introduction
0:40 Domain
2:25 Solving Quartic equation
2:45 Algebraic manipulations
5:44 Algebraic identities
8:49 Solving quadratic equations
9:58 Quadratic formula
11:15 Solutions

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Thanks for Watching!
@infyGyan

[gkwugqbfigvjg]

x=racine carré de(6)
ou
x=3+racine carré de(15)

[shmuelzehavi]

Nice solution.
As a matter of fact, you restricted yourself not only to real x values but also to real functions of x inside the original equation. Without such a restriction, x = - √6 and x = 3 - √15 would be as well real valued roots satisfying the original equation, whose internal functions would become complex valued ( √x = i ∜6 and √x = 0.934 i ) .

[woobjun]

why don't you multiply directly the given as (1st x 4th) and (2nd x 3rd)? It yield (t - 5)(t - 1)=0 where t = x - (6/x) to solve for t then x

[walterwen]

Crack the Challenge: (√x + 1/√x)(√x + 2/√x)(√x – 3/√x)(√x – 6/√x) = 5; x = ?
x > 0, Real value roots; x < 0, Imaginary/complex roots, if acceptable
(√x + 1/√x)(√x + 2/√x)(√x – 3/√x)(√x – 6/√x) = [(x + 1)(x + 2)(x – 3)(x – 6)]/x²
= [(x + 1)(x – 6)][(x + 2)(x – 3)]/x² = [(x² – 6 – 5x)(x² – 6 – x)]/x² = 5
(x² – 6)² – 6x(x² – 6) + 5x² = 5x²; (x² – 6)² – 6x(x² – 6) = 0, (x² – 6)(x² – 6 – 6x) = 0
x² – 6 = 0, x² = 6; x = ± √6 or x² – 6x – 6 = 0, (x – 3)² = (√15)²; x = 3 ± √15
Answer check:
(√x + 1/√x)(√x + 2/√x)(√x – 3/√x)(√x – 6/√x) = [(x² – 6 – 5x)(x² – 6 – x)]/x²
x = ± √6; x² – 6 = 0, [(– 5x)(– x)]/x² = (5x²)/x² = 5; Confirmed
x = 3 ± √15; x² – 6 = 6x, [(6x – 5x)(6x – x)]/x² = (5x²)/x² = 5; Confirmed
Final answer:
Real value roots; x = √6 or x = 3 + √15
Imaginary/complex roots; x = – √6, or x = 3 – √15

[ericerpelding]

I like this problem.
Here is my solution.
(sqrt(x) + 1/sqrt(x))(sqrt(X) + 2/sqrt(x))(sqrt(X) - 3/sqrt(x))(sqrt(X) - 6/sqrt(x)) = 5.
Rearrange.
(sqrt(x) + 2/sqrt(x))(sqrt(X) - 3/sqrt(x))(sqrt(X) + 1/sqrt(x))(sqrt(X) - 6/sqrt(x)) = 5.
Multiply the first two factors together, and the last two factors together.
(x - 6/x - 1)(x - 6/x -5) = 5.
Let t = x - 6/x, then (t - 1)(t - 5) = 5, which gives t^2 - 6t + 5 = 5.
Subtract 5 from both sides to give t(t - 6) = 0.
This has solutions t = 0 and t = 6.
If t = 0, then x - 6/x = 0, which has the positive solution x = sqrt(6).
If t = 6, then x - 6/x = 6, which gives x^2 -6x - 6 = 0, a quadratic equation with a positive solution x = 3 + sqrt(15).
The solutions for x are:
x = sqrt(6) and x = 3 + sqrt(15).