Problem 53 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The real solution is \( x = 2 \).

Step 1: Understand the Problem

We need to find all real values of \( x \) such that the equation \( \sqrt{x + \sqrt{x + 2}} = 2 \) holds true. This involves isolating \( x \) by manipulating the equation to remove the square roots.

Step 2: Remove the Outer Square Root

Square both sides of the equation to eliminate the outer square root. This leads to:\[(\sqrt{x + \sqrt{x + 2}})^2 = 2^2 \x + \sqrt{x + 2} = 4\]

Step 3: Isolate the Inner Square Root

Subtract \( x \) from both sides to isolate the inner square root:\[\sqrt{x + 2} = 4 - x\]

Step 4: Remove the Inner Square Root

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

Step 5: Bring Everything to One Side

Rearrange the terms to bring all terms to one side of the equation:\[x^2 - 9x + 14 = 0\]

Step 6: Solve the Quadratic Equation

Factor the quadratic equation to find the values of \( x \):\[(x - 7)(x - 2) = 0\]The solutions are \( x = 7 \) and \( x = 2 \).

Step 7: Validate the Solutions

Check each solution in the original equation:**For \( x = 7 \):**\[\sqrt{7 + \sqrt{7 + 2}} = \sqrt{7 + 3} = \sqrt{10} eq 2\]This is not a valid solution.**For \( x = 2 \):**\[\sqrt{2 + \sqrt{2 + 2}} = \sqrt{2 + 2} = \sqrt{4} = 2\]This is a valid solution.

Key Concepts

Real Solutions

Real solutions refer to the values of variables that satisfy an equation without involving imaginary numbers. In our exercise, we need to find all real solutions for the equation \( \sqrt{x + \sqrt{x + 2}} = 2 \). The process of finding these solutions involves transforming the equation to identify any real numbers that make it true.

As illustrated in the solution steps, we simplified the given equation through various operations, such as squaring and isolating terms. This helps to avoid complex numbers, ensuring our solution set only contains real numbers. For this specific equation, the real solution is \( x = 2 \), as it satisfies the original equation perfectly.

Square Roots

Square roots are essential in algebra. A square root of a number \( y \) is a number \( x \) such that \( x^2 = y \). In our equation, both the outer and the inner expressions are under square root symbols.

To solve \( \sqrt{x + \sqrt{x + 2}} = 2 \), we begin by squaring both sides to remove the outer square root. This converts the equation into one without the initial square root, making it simpler to manipulate.

• Start by squaring: \( x + \sqrt{x + 2} = 4 \)
• Proceed to isolate and remove the second square root: \( \sqrt{x+2} = 4 - x \)

These steps ensure that we handle the square roots correctly, and preparation for validation of the solution.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). When transformed, our equation becomes \( x^2 - 9x + 14 = 0 \).

To solve a quadratic equation, we can factor it, use the quadratic formula, or complete the square. Here, it was factored as \((x-7)(x-2) = 0 \), yielding potential solutions of \( x = 7 \) and \( x = 2 \).

• Factoring allows us to break down the equation into simpler, solvable parts.
• Quadratics often yield two potential solutions, which we must verify.

Understanding how to solve quadratic equations is crucial in identifying real solutions.

Validation of Solutions

After finding potential solutions to an algebraic equation, it's vital to validate them. Validation ensures that each solution satisfies the original equation. For our problem, we found \( x = 7 \) and \( x = 2 \) as possible solutions from the quadratic equation.

Validation Steps:

• Substitute back into the original equation to check correctness.
• For \( x = 7 \), the equation \( \sqrt{7 + 3} = \sqrt{10} eq 2 \) shows it's not valid.
• For \( x = 2 \), \( \sqrt{2+2} = \sqrt{4} = 2 \) confirms this is a valid solution.

Thorough verification prevents incorrect solutions, confirming only valid results are considered.