Problem 134 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the value of \(\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}\)

Step-by-Step Solution

Answer

The value of the infinite nested radical \(\sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}\) is \(x = 2\).

Step 1: Assign a variable to the expression

Let x be the value of the infinite nested radical: \(x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \ldots}}}\)

Step 2: Simplify the radical expression

Since x is the entire infinite nested radical, it also includes the entire radical expression inside the square root: \(x = \sqrt{2 + x}\)

Step 3: Square both sides of the equation

To get rid of the square root, we'll square both sides of the equation: \(x^2 = 2 + x\)

Step 4: Rearrange the equation

Rearrange the equation to solve for x: \(x^2 - x - 2 = 0\)

Step 5: Solve the quadratic equation

Use the Quadratic Formula to solve the equation: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Here, a = 1, b = -1, and c = -2. Plug these values into the Quadratic Formula to find x: \(x = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}\)

Step 6: Simplify the expression

After plugging in the values, simplify the expression to find the two possible values of x: \(x = \frac{1 \pm \sqrt{9}}{2}\) \(x = \frac{1 \pm 3}{2}\) So, x can be either -1 or 2.

Step 7: Choose the valid value of x

Only the positive value makes sense in this context (a negative square root doesn't have a real value), so we choose x = 2 as the valid value. So, the value of the infinite nested radical is \(x = 2\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in understanding many mathematical concepts, including the solution of nested radicals like the one in our example. These equations take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The solution to a quadratic equation provides the values for \(x\) that make the equation true.
To solve these equations, you can use several methods such as factoring, completing the square, or applying the quadratic formula. In our problem, we used the quadratic formula:

1. First, rearrange the equation to standard form.
2. Download the coefficients: \(a\), \(b\), and \(c\).
3. Plug these into the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Using the quadratic formula is often the most efficient way to solve when factoring is not feasible. This step is critical as it transforms a complex nested radical into a more manageable form.

Infinite Series

Infinite series appear in many branches of mathematics and can describe complex phenomena with simple representations. An infinite series is essentially a summation of an infinite list of numbers which, interestingly, can converge to a finite value.
In the context of our problem, each layer of the radical can be thought of as adding another term to an infinite series. Here, the nested radical \(\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}\) can be imagined as continuing infinitely, similar to an infinite series.
The key insight is that despite its complexity and infinite nature, the expression has a stable value. By assuming a constant solution exists for the nested radicals and solving for it using algebra, we find the converging result. In this case, the infinite series stabilizes at 2. By solving such problems, we not only harness the infinite but create a bridge of understanding between imagination and reality.

Radical Expressions

Radical expressions involve roots, such as square roots, cube roots, etc. They can seem daunting, especially when nested, like in our exercise. A nested radical is a radical expression that contains another radical expression within itself, creating layers of complexity. In \(\sqrt{2+\sqrt{2+\sqrt{2+\ldots}}}\), each inside layer repeats the process, forming an infinite loop.
To simplify a nested radical, it's necessary to assume it equates to a fixed number, which allows us to algebraically manipulate the expression into a form that can be solved. With nested radicals, concepts like logarithms and exponentials often come into play.
The process involves isolating one layer of the radical to simplify and solve the equation step by step. For example, we rewrote it as \(x = \sqrt{2 + x}\), squaring both sides to solve the quadratic equation that emerged. Such techniques provide powerful tools to understand and work with complex radical expressions effectively.