Surds and Indices Part 4
Duration: 3 min
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This educational video, presented by Yash Jain Sir from Knowledge Gate Educator, is a tutorial on solving problems involving infinite nested radicals, specifically focusing on surds and indices. The video begins with a worked example demonstrating the simplification of the square root of 2601 by factoring it into prime factors (3x3x17x17), which is then simplified to 3x17, resulting in 51. The main lesson then transitions to a more complex problem: finding the value of the infinite nested radical y = √(3√(9√(3√(9...)))) where the pattern '3, 9' repeats infinitely. The instructor explains that this expression can be rewritten as y = √(3√(3²√(3√(3²...)))). By recognizing the repeating pattern, he sets up the equation y = √(3√(3²y)), which is then squared to eliminate the outermost radical, leading to y² = 3√(3²y). This is further simplified to y² = 3√(9y), and squaring both sides again yields y⁴ = 9 * 9y, or y⁴ = 81y. Dividing both sides by y (assuming y ≠ 0) gives y³ = 81. Since 81 is 3⁴, the equation becomes y³ = 3⁴. The instructor then solves for y by taking the cube root of both sides, resulting in y = 3^(4/3). The video concludes with a similar example using the number 2, where y = √(2√(4√(2√(4...)))) is solved to find y = 2^(4/3). The final frame is a 'Thanks for Watching' screen.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a whiteboard displaying a math problem: √2601 = √(3x3x17x17), which is simplified to 3x17, resulting in 51. The topic is introduced as 'SURDS & INDICES'. The instructor, Yash Jain Sir, then presents a new problem: y = √(3√(9√(3√(9...)))) with the pattern '3, 9' repeating infinitely. He explains that this can be rewritten as y = √(3√(3²√(3√(3²...)))). He sets up the equation y = √(3√(3²y)) and begins to solve it by squaring both sides, leading to y² = 3√(3²y). He then simplifies this to y² = 3√(9y) and squares both sides again to get y⁴ = 9 * 9y, or y⁴ = 81y. He divides by y to get y³ = 81, and since 81 is 3⁴, he writes y³ = 3⁴. He then circles the number 3 and writes '3' below it, indicating the next step is to solve for y.
2:00 – 2:42 02:00-02:42
The instructor continues solving the equation y³ = 3⁴. He explains that to find y, one must take the cube root of both sides, so y = (3⁴)^(1/3) = 3^(4/3). He then presents a similar problem with the number 2: y = √(2√(4√(2√(4...)))) and applies the same method. He rewrites it as y = √(2√(2²√(2√(2²...)))) and sets up the equation y = √(2√(2²y)). He squares both sides to get y² = 2√(2²y), simplifies to y² = 2√(4y), and squares again to get y⁴ = 4 * 4y, or y⁴ = 16y. Dividing by y gives y³ = 16. Since 16 is 2⁴, he writes y³ = 2⁴. He then circles the number 2 and writes '2' below it, indicating the solution is y = 2^(4/3). The video ends with a 'Thanks for Watching' screen.
The video provides a clear, step-by-step demonstration of solving infinite nested radical expressions by leveraging the properties of exponents and indices. The core method involves recognizing the repeating pattern within the radical, which allows the entire expression to be represented as an equation in terms of the variable itself. This self-referential equation is then algebraically manipulated by squaring both sides to eliminate radicals, leading to a polynomial equation that can be solved for the variable. The lesson effectively connects the concepts of surds and indices, showing how to convert radicals into fractional exponents to simplify complex expressions. The progression from a simple numerical example to a more abstract algebraic one reinforces the general applicability of the technique.