Mastering An Engaging Rational Equation Challenge | Algebra

Mastering An Engaging Rational Equation Challenge | Algebra

Welcome to our latest algebra tutorial where we delve into mastering rational equations! In this engaging video, we'll tackle challenging rational equations step-by-step, providing you with the tools and strategies needed to conquer even the most complex problems. From understanding the fundamentals to mastering advanced techniques, this tutorial is designed to enhance your algebra skills and boost your problem-solving confidence. Whether you're a student looking to ace your next math exam or someone who wants to brush up on algebraic concepts, this video is for you. Get ready to level up your algebra game and become a master of rational equations!

🔮 In this video, you'll discover:

Method to confidently solve rational equation challenge.
Step-by-step guidance on applying magical technique to simplify and solve.
Pro tips for mastering the art of rational equation mastery.
Get ready to command and conquer with this method! 🚀✨ Subscribe now for a journey into the realm of mathematical mastery!

Timestamps:
0:00 Introduction
0:27 Domain
1:20 Algebraic identity
4:28 Substitution
6:27 Quadratic equation
7:52 Quadratic formula
8:20 Solutions
8:45 Verification

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Thanks for Watching !!
@infyGyan ​

[kassuskassus]

After some transformations, we'll get (x^2/X-2)^2-4(x^2/x-2)=5. Let t=x^2/x-2. We get t^2-4t-5=0. We solve for t (t=-1 and t=5). Then, we solve for x. The two real solutions are x=-2, x=1 and the two others are complex x=(5-iroot15)/2 and x=(5+iroot15)/2.

[woobjun]

Thanks a lot.
As another try, direct expantion yields x^4 - 4x^3 +3x^2 +20x -20 = 0, which gives a solution x=1 with x^3 -3x^2 +20 =0 by SDM.
To solve (x^3 + 8) - 3(x^2 - 4) = 0, results in (x+2)(x^2 -5x +10) = 0.
Thus x=1, -2 and 2 CSs.

[NadiehFan]

The equation to solve is

x² + (2x/(x − 2))² = 5

This equation can be solved elegantly (1) without having to mess with fractions, (2) without substitutions, and (3) without ending up with unelegant expressions for the roots which still need to have their denominators rationalized. The key here is to use a technique which I demonstrated earlier, i.e. converting a product of quantities into a difference of squares.

If we start by multiplying both sides by (x − 2)² to eliminate the fraction (which is allowed since x ≠ 2) we get

x²(x − 2)² + 4x² = 5(x − 2)²

I ultimately want to create squares on both sides, and to do so I first incorporate the term 4x² at the left hand side into the product x²(x − 2)² which can be done since they both have a common factor x². Expanding (x − 2)² at the left hand side we have

x²(x² − 4x + 4) + 4x² = 5(x − 2)²

and taking out the common factor x² at the left hand side this gives

x²((x² − 4x + 4) + 4) = 5(x − 2)²

which is

x²(x² − 4x + 8) = 5(x − 2)²

We now have a product of two quantities x² and x² − 4x + 8 at the left hand side which we can turn into a difference of squares. The average (arithmetic mean) of x² and x² − 4x + 8 is half their sum which is x² − 2x + 4 and the difference between x² and x² − 4x + 8 is x² − (x² − 4x + 8) = 4x − 8 = 4(x − 2) so half their difference is 2(x − 2). So, we have x² = (x² − 2x + 4) + 2(x − 2) and x² − 4x + 8 = (x² − 2x + 4) − 2(x − 2) and we can therefore rewrite the equation as

((x² − 2x + 4) + 2(x − 2))((x² − 2x + 4) − 2(x − 2)) = 5(x − 2)²

and applying the difference of two squares identity (a + b)(a − b) = a² − b² to the left hand side this can be written as

(x² − 2x + 4)² − 4(x − 2)² = 5(x − 2)²

and bringing the term 4(x − 2)² from the left hand side over to the right hand side this gives

(x² − 2x + 4)² = 9(x − 2)²

We now have a square on both sides of our equation because 9(x − 2)² = 3²(x − 2)² = (3(x − 2))² = (3x − 6)² so we have

(x² − 2x + 4)² = (3x − 6)²

From here we can proceed in two ways which are fundamentally the same. One way is to bring the square from the right hand side over to the left hand side to create a difference of two squares on the left hand side whereas the right hand side becomes zero. We can then factor the left hand side into two quadratic polynomials using the difference of two squares identity a² − b² = (a − b)(a + b) and apply the zero product property which says that a product is zero if and only if at least one of its factors is itself zero. The other way is to use the equivalence

A² = B² ⟺ A = B ⋁ A = −B

In English: if the squares of two quantities are equal, then these quantities are either equal _or_ each others opposite, and vice versa. The two ways are fundamentally the same because this equivalence is a simple consequence of the difference of two squares identity and the zero product property, since we have

A² = B² ⟺ A² − B² = 0 ⟺ (A − B)(A + B) = 0 ⟺ A − B = 0 ⋁ A + B = 0 ⟺ A = B ⋁ A = −B

Now, returning to our equation and applying the equivalence A² = B² ⟺ A = B ⋁ A = −B we have

x² − 2x + 4 = 3x − 6 ⋁ x² − 2x + 4 = −3x + 6

which gives

x² − 5x + 10 = 0 ⋁ x² + x − 2 = 0

and now we only need to solve these two quadratic equations to obtain all four roots of our equation. The second quadratic x² + x − 2 = 0 can easily be factored because we have

(x − 1)(x + 2) = 0

which gives

x = 1 ⋁ x = −2

The first quadratic x² − 5x + 10 = 0 cannot be factored over the integers and its discriminant is Δ = (−5)² − 4·1·10 = 25 − 40 = −15, so, using the quadratic formula, this quadratic gives two complex solutions

x = (5 + i√15)/2 ⋁ x = (5 − i√15)/2

which completes the solution of the equation.

[SrisailamNavuluri]

It is in the form a^2+b^2=5
(a+b)^2-2ab=5
Now simplifying and solving
x^2-5x+10=0
x^2+x-2=0
x=1 or -2 or 2 complex roots.

[alexandrovich]

These transformations are not obvious. I dislike math problems where you don't measures skill or knowledge, but rather pulling a trick off the sleeve. It seems more of a magic act than math solving abilities...