Problem 54 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ \sqrt{1+\sqrt{x+\sqrt{2 x+1}}}=\sqrt{5+\sqrt{x}} $$

Step-by-Step Solution

Answer

The equation is complex; simplify and check roots manually may help find solutions.

Step 1: Square Both Sides

Start by squaring both sides of the equation to eliminate the outer square roots. We have:\[ ( ext{LHS})^2 = ( ext{RHS})^2 \]\[ (1 + \sqrt{x + \sqrt{2x + 1}}) = (5 + \sqrt{x}) \]

Step 2: Isolate the Inner Nested Root

Rearrange the equation to further isolate the nested square root:\[ \sqrt{x + \sqrt{2x + 1}} = 4 + \sqrt{x} \]

Step 3: Square Again to Simplify

Square both sides again to eliminate the nested square root:\[ x + \sqrt{2x + 1} = (4 + \sqrt{x})^2 \]\[ x + \sqrt{2x + 1} = 16 + 8\sqrt{x} + x \]

Step 4: Simplify and Isolate the Remaining Root

Now simplify and isolate the remaining square root:\[ \sqrt{2x + 1} = 16 + 8\sqrt{x} - x \]Cancel out the \(x\) terms and simplify further if possible.

Step 5: Simplify and Solve

At this stage, check if further squaring or simplifying can help isolate \(x\). Unfortunately, doing so again makes the equation very complex, so try another approach to estimate or directly solve for solutions.

Key Concepts

Nested Square Roots

Nested square roots occur when there are square roots within square roots. This is a common scenario in certain algebraic equations and can often make solving them a bit tricky. In the equation we begin with, we have more than one layer of square roots to deal with. These can be found on both sides of the equation:

• Left-hand side: \(\sqrt{1+\sqrt{x+\sqrt{2x+1}}}\)
• Right-hand side: \(\sqrt{5+\sqrt{x}}\)

To tackle these, we need to systematically work our way from the outside to the inside. By doing this, we progressively remove the layers until all the square roots are eliminated. This unraveling often requires a methodical squaring of each side of the equation in a step-by-step process.

Solving Equations

Solving equations, especially with nested square roots, involves systematic steps to simplify and isolate variables. Here's a general approach:

• First, square both sides of the equation to remove the outermost square roots.
• Next, rearrange and simplify the equations step by step. Squaring further may be necessary to eliminate remaining nested roots.

In our step-by-step process, we initially square both sides of the equation to begin simplifying the nested structure. Through simplifying and re-squaring steps, we peel away each layer of square root to expose the basic equation underneath. When we square, it's crucial to simplify and check if further squaring is needed or if another method should be applied.

Real Solutions

When searching for real solutions to equations, especially those involving square roots, it's essential to ensure the solution is within the realm of real numbers. Square roots are only defined for non-negative numbers in the set of real numbers.

• This means any terms under a square root must be positive to have a real solution.
• Checking for extraneous roots is critical. This happens when steps like squaring both sides introduce solutions that don't satisfy the original equation.

After we simplify and attempt to solve the equation, we must verify solutions by substituting back into the original equation to confirm they hold true. Real solutions will satisfy the structure and constraints set by the equation's initial parameters.

Transcendental Functions

Transcendental functions aren't present in the given equation itself, but understanding their distinction is helpful. Transcendental functions, unlike algebraic functions, cannot be expressed using a finite number of algebraic operations (addition, multiplication, etc.). Examples include:

• Exponential functions \(e^x\)
• Trigonometric functions like \(\sin x\)
• Logarithmic functions like \(\ln x\)

Recognizing whether a function is transcendental or algebraic is important in algebra problem solving. While the problem at hand focuses on algebraic operations and real solutions, familiarity with transcendental functions expands one's understanding of function types and their properties. This broader comprehension aids in better tackling complex equations that might involve such functions in more advanced mathematical contexts.