Can You Solve This Intriguing Radical Equation?

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Can You Solve This Intriguing Radical Equation?
Prepare to put your problem-solving skills to the test with this intriguing radical equation! 🧩🔍 In this video, we'll dive into the depths of a complex equation and guide you through the process of solving it step by step. Whether you're a math enthusiast seeking a challenge or a student preparing for exams, this equation will surely captivate your interest. Join us as we unravel the mystery and uncover the solution together! Don't forget to like, comment, and subscribe for more fascinating math content. 📐✨
Topics covered:
Algebra
Radical Equation
Substitution
Algebraic identities
Algebraic manipulations
Quadratic equations
Quadratic formula
Synthetic division method
Rational root theorem
Math Tutorial
Math Olympiad Preparation
Timestamps:
0:00 Introduction
1:16 Substitution
3:00 Quartic equation
4:25 SDM
5:35 Rational root theorem
8:30 Solutions
9:38 Verification
#radicalequations #mathpuzzle #problemsolving #mathchallenge #equationsolving #mathtutorial #education #stem #mathematics #algebra #matholympiad
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@infyGyan

[@docnelson2008]

I get the same solutions by squaring both sides and solving (after rearranging the terms) the resultant polynomial x^4+8x^3+9x^2+18x+10=0. Clearly (x-1) is a factor so after division we're left with x^3+9x^2+8x+10=0. (x+1) is a factor so after further division we have x^2+8x+10=0 with solutions x=-4+-SQRT(6). Remaining solutions are, of course, x=1, x=-1

[@NadiehFan]

My solution, before being able to watch the video:
The equation to solve is
(x + 2)² = √(15x² + 40x + 26)
Square both sides and we get
(x + 2)⁴ = 15x² + 40x + 26
x⁴ + 8x³ + 24x² + 32x + 16 = 15x² + 40x + 26
x⁴ + 8x³ + 9x² − 8x − 10 = 0
x⁴ + 8x(x² − 1) + 9x² − 10 = 0
x²(x² − 1) + 8x(x² − 1) + 10x² − 10 = 0
x²(x² − 1) + 8x(x² − 1) + 10(x² − 1) = 0
(x² − 1)(x² + 8x + 10) = 0
(x − 1)(x + 1)((x + 4)² − 6) = 0
x − 1 = 0 ⋁ x + 1 = 0 ⋁ (x + 4)² − 6 = 0
x = 1 ⋁ x = −1 ⋁ x = −4 + √6 ⋁ x = −4 − √6
So we have four real solutions, but now we still need to verify these solutions because the radicand 15x² + 40x + 26 at the right hand side of the original equation must not be negative. It is easy to see that this condition is satisfied for x = 1 and x = −1, but what about the irrational solutions x = −4 + √6 and x = −4 − √6 which are both negative?
Note that you are _not allowed to use a calculator_ when solving problems that appeared on math contests. We can proceed as follows. Note that x = −4 + √6 and x = −4 − √6 are both solutions of the quadratic x² + 8x + 10 = 0 so for each of these values of x we have
x² = −8x − 10
and therefore
15x² + 40x + 26 = 15(−8x − 10) + 40x + 26 = −80x − 124
So, for x = −4 − √6 we have
15x² + 40x + 26 = −80x − 124 = −80(−4 − √6) − 124 = 196 + 80√6
which is evidently positive. But for x = −4 + √6 we have
15x² + 40x + 26 = −80x − 124 = −80(−4 + √6) − 124 = 196 − 80√6
and it is not immediately obvious whether this is positive or negative. So how do we decide this _without_ using any calculating device? Well, we have
(196 + 80√6)(196 − 80√6) = 196² − (80√6)²
= (200 − 4)² − 6400·6
= 40000 − 1600 + 16 − 36000 − 2400
= 40000 − 36000 − 4000 + 16
= 16
So, since 196 + 80√6 is positive and since the product of 196 + 80√6 and 196 − 80√6 is positive, this means that 196 − 80√6 itself must also be positive. So, for x = −4 + √6 the radicand 15x² + 40x + 26 = 196 − 80√6 is indeed positive. Therefore, all four solutions
x = 1 ⋁ x = −1 ⋁ x = −4 + √6 ⋁ x = −4 − √6
are accepted as solutions of the equation (x + 2)² = √(15x² + 40x + 26).

[@StaR-uw3dc]

Excellent explanation.
The only comment: we can notice that -80x-124 = 4(-20x-31) so it is sufficient to verify whether -20(-4-√6)-31= 49+20√6 and -20(-4+√6)-31= 49-20√6 is nonnegative. These numbers are reciprocals and the first one is positive so the second one is also positive.

[@ZhilinChen-my7tp]

我也覺得四個答案才對。

[@ricardonarcizovillarroelvi1786]

En algunos casos es más fácil y rápido aplicar la solución particular de la ecuación cuadrática con coeficiente para x^2 de 1, cuya formula es:
X=-(b/2)+_sqrt((b/2)^2-c)
Dónde a=1