A Rational Challenge: Unlocking the Extraordinary Solution

A Rational Challenge: Unlocking the Extraordinary Solution

🔍 Embark on a mathematical journey with us in "A Rational Challenge: Unlocking the Extraordinary Solution"! 🧠💡 Dive into the world of rational equations as we tackle a unique challenge and unveil an extraordinary solution that will redefine the way you approach problem-solving. 🚀✨

🔗 Join us as we break down the complexity, providing step-by-step insights into the method behind unlocking this exceptional solution. Whether you're a student, educator, or math enthusiast, this video is designed to elevate your understanding and appreciation for rational equations. 🎓🔢

Topics covered:
Algebra Challenge
Rational Equation
Quartic equation
How to solve Rational Equation?
Quadratic equations
Math Olympiad
Algebra
Math Tricks
Algebraic identities
Sum of roots
Algebraic manipulations
Substitutions
Real Solutions
Quadratic formula
Math Tutorial
Algebraic Challenging Equations
Math Olympiad Preparation

Timestamps:
0:00 Introduction
0:27 Method-1
0:43 Substitution
1:50 Sum of roots
4:05 Quadratic equations
8:41 Solutions
8:50 Method-2
10:18 Quartic equation
11:35 Synthetic division
15:29 Solutions
15:40 Verification

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We'd love to hear from you! Did you manage to solve the equation? What other math problems would you like us to cover? Let us know in the comments below!

🎓 Happy learning, and see you in the next video! 🎉

Thanks for Watching !!
@infyGyan ​

[kassuskassus]

Thank you very much sir ! I used the second method.

[NadiehFan]

I started out exactly as you do in your first method, but I found the solutions in simpler way. The equation to solve is

(1) x²((17 − x)/(x + 1)) + x((17 − x)/(x + 1))² = 72

Subtituting

(2) (17 − x)/(x + 1) = y

the equation to solve becomes

x²y + xy² = 72

which we can write as

(3) xy(x + y) = 72

But multiplying both sides of (2) by x + 1 (which is fine since x ≠ −1) we obtain

17 − x = xy + y

which gives

(4) x + y = 17 − xy

and substituting (4) in (3) we get

xy(17 − xy) = 72
17xy − (xy)² = 72
(xy)² − 17xy + 72 = 0
(xy − 8)(xy − 9) = 0
xy = 8 ⋁ xy = 9

So, the product xy is either 8 or 9. Substituting xy = 8 in (3) gives x + y = 9 and substituting xy = 9 in (3) gives x + y = 8 so we have

(xy = 8 ⋀ x + y = 9) ⋁ (xy = 9 ⋀ x + y = 8)
(xy = 8 ⋀ y = 9 − x) ⋁ (xy = 9 ⋀ y = 8 − x)
x(9 − x) = 8 ⋁ x(8 − x) = 9
x² − 9x + 8 = 0 ⋁ x² − 8x + 9 = 0

which gives the solutions

x = 1 ⋁ x = 8 ⋁ x = 4 + √7 ⋁ x = 4 − √7

Of course, instead of eliminating y we could also have used Vieta's formulas for the product and sum of the roots of a quadratic equation. The first possibility (xy = 8 ⋀ x + y = 9) means that x is either of the roots of the quadratic equation t² − 9t + 8 = 0 so x = 1 or x = 8 and the second possibility (xy = 9 ⋀ x + y = 8) means that x is either of the roots of the quadratic equation t² − 8t + 9 = 0 so x = 4 + √7 or x = 4 − √7. For each of the two possibilities, y will be of course be the other root of the same quadratic equation.

[ndthhyl]

X=1 est evdent. On peut developer et diviser.