COMO RESOLVER EQUAÇÕES IRRACIONAIS COMPLEXAS SEM ERRO?
Matemática com Cristiano Marcell
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In this video, we will solve step by step... I'm stepping into a classic irrational equation that mixes square roots and cube roots: $\sqrt{x+1} - \sqrt[3]{2x-6} = 2$.
The big secret to not getting lost in the calculations is to use a convenient change of variables ($a$ and $b$), apply the notable product of the cube of the difference, and factor the resulting equation to find all possible values of X.
📐 Topics covered in this lesson: • Change of variables in irrational equations • Elimination of radicals (square and cube) • Development of the cube of the difference $(a - b)^3$ • Factoring and factoring out terms • Solving the quadratic equation (Bhaskara / Sum and Product) • Determining the solution set for $x$
⏱️ Chapters: 00:00 - Introduction to the irrational equation 00:32 - Change of variable and isolation of terms 02:08 - Cubing and setting up the system 04:45 - Applying the notable product of the cube of the difference 07:43 - Factoring and solving the quadratic equation 10:48 - Finding the final solutions for X 12:48 - Solution set and closure
👍 Did you like the solution? Leave your like, subscribe to the Mathematics with Cristiano Marcell channel and share with those who love a good math challenge!
#Mathematics #Algebra #IrrationalEquation #NotableProducts #BasicMathematics
20 Comments
Professor não erra, se engana.
Esta soma coincidiu com a igualdade 5-x. No final teve 2 resultados pra x. Um deles era -1/2
Gostaria de rever essa resolução. Abraço.
a - 2 = b e 2a² - 8 = b³, com a ≥ 0 e b ∈ ℝ. Daí,
2(a² - 4) = b³
2(a - 2)(a + 2) = b³
2b(a + 2) = b³. Note que b = 0 é uma solução. Se b ≠ 0, então:
2(a + 2) = b²
2(a - 2 + 4) = b²
2(b + 4) = b²
b² - 2b - 8 = 0
(b - 1)² - 9 = 0
(b - 1)² = 9
b - 1 = ±√9 = ±3
b = 4 ou b = -2. Assim,
● b = 0 ⟹ 2x - 6 = 0 ⟹ x = 3.
● b = 4 ⟹ 2x - 6 = 4³ ⟹ x = 35.
● b = -2 ⟹ 2x - 6 = (-2)³ ⟹ x = -1.
∛︎(2x-6) = y ⇒︎ x=(y³+6)/2
[√︎(x+1) = y+2]² ⇒︎ x+1=(y+2)²
(y³+6)/2 + 1 = y²+4y+4
y³-2y²-8y=0
∴︎ y=0 ⇒︎ x=3
y=4 ⇒︎ x=35
y=-2 ⇒︎ x=-1
y = √(x + 1)
2y^2 - 8 = y^3 - 6y^2 + 12y - 8
y^3 - 8y^2 + 12y = 0
y(y^2 - 8y + 12) = 0
y(y - 2)(y - 6) = 0
√(x + 1) = y = 0, 2, 6
(x + 1) = 0, 4, 36
x = -1, 3, 35 ✅
COMO RESOLVER EQUAÇÕES IRRACIONAIS COMPLEXAS SEM ERRO?
Matemática com Cristiano Marcell
8,767 views
📚 Link to my Basic Math course 👍 Did you enjoy the lesson? Subscribe to the Math with Cristiano Marcell channel, leave a like, and share with anyone who likes algebra and math challenges!
📚 STUDY WITH ME
👉 https://pay.kiwify.com.br/9E7BSPy
📚 Link to purchase my book
👉 https://www.editorarideel.com.br/os-1...
🇧🇷 Use my 15% discount coupon 👇👇👇👇👇👇👇👇👇👇 CRISTIANO31
📚 Link to my course PASS FOR PROFMAT https://pay.kiwify.com.br/vVxJCPz
📲 SOCIAL MEDIA: • Instagram: @prof.cristianomarcell
Business contact: mat_amorim@hotmail.com
In this video, we will solve step by step... I'm stepping into a classic irrational equation that mixes square roots and cube roots: $\sqrt{x+1} - \sqrt[3]{2x-6} = 2$.
The big secret to not getting lost in the calculations is to use a convenient change of variables ($a$ and $b$), apply the notable product of the cube of the difference, and factor the resulting equation to find all possible values of X.
📐 Topics covered in this lesson: • Change of variables in irrational equations • Elimination of radicals (square and cube) • Development of the cube of the difference $(a - b)^3$ • Factoring and factoring out terms • Solving the quadratic equation (Bhaskara / Sum and Product) • Determining the solution set for $x$
⏱️ Chapters: 00:00 - Introduction to the irrational equation 00:32 - Change of variable and isolation of terms 02:08 - Cubing and setting up the system 04:45 - Applying the notable product of the cube of the difference 07:43 - Factoring and solving the quadratic equation 10:48 - Finding the final solutions for X 12:48 - Solution set and closure
👍 Did you like the solution? Leave your like, subscribe to the Mathematics with Cristiano Marcell channel and share with those who love a good math challenge!
#Mathematics #Algebra #IrrationalEquation #NotableProducts #BasicMathematics