Problem 59 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \sqrt{x+\sqrt{x+2}}=2 $$

Step-by-Step Solution

Answer

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

Step 1: Isolate the Inner Square Root

Start by squaring both sides of the equation to eliminate the outer square root. The original equation is \( \sqrt{x+\sqrt{x+2}} = 2 \). Square both sides to get: \[ x + \sqrt{x+2} = 4.\]

Step 2: Isolate the Remaining Square Root

Next, isolate \( \sqrt{x+2} \) by subtracting \( x \) from both sides of the equation: \[ \sqrt{x+2} = 4 - x.\]

Step 3: Eliminate the Square Root

Square both sides again to remove the square root. This results in: \[ x+2 = (4-x)^2.\]

Step 4: Expand and Rearrange the Equation

Expand \( (4-x)^2 \) which gives \( 16 - 8x + x^2 \). Substitute this back into the equation: \[ x + 2 = x^2 - 8x + 16.\] Rearrange the terms to form a quadratic equation: \[ x^2 - 9x + 14 = 0.\]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \( x^2 - 9x + 14 = 0 \) using factoring or the quadratic formula. Factoring gives: \[ (x - 7)(x - 2) = 0.\]This implies \( x = 2 \) or \( x = 7 \).

Step 6: Verify Potential Solutions

Ensure both solutions \( x = 2 \) and \( x = 7 \) satisfy the original equation: For \( x = 2 \): \[ \sqrt{2+\sqrt{2+2}} = \sqrt{2+2} = \sqrt{4} = 2.\]For \( x = 7 \): \[ \sqrt{7+\sqrt{7+2}} = \sqrt{7+3} = \sqrt{10} eq 2.\]Thus, \( x=7 \) is not a solution.

Key Concepts

Real solutions

In algebra problems, particularly with equations that include roots or quadratic expressions, we're often asked to find 'real solutions.' A real solution refers to any solution that is a real number. Essentially, these are numbers you can find on the number line. Contrast this with imaginary solutions, which involve the imaginary unit \( i \) (where \( i^2 = -1 \)), and don't appear on the ordinary number line, but are on the complex plane instead.

When solving equations like \( \sqrt{x+\sqrt{x+2}} = 2 \), we're particularly interested in real solutions because these reflect practical values which can be used in applications that correspond to real-world quantities. Here's how you find them:

• Carefully simplify the expression to eliminate roots.
• Convert it into an easily solvable form, like a quadratic equation if needed.
• Verify that any solutions you find satisfy the original equation without resulting in contradictions.

By confirming that the solution truly works when substituted back into the original equation, you ensure it is a valid real solution.

Quadratic equation

A quadratic equation is a polynomial equation of the second degree, typically in the form \( ax^2 + bx + c = 0 \). The general characteristic of a quadratic equation is that one of the terms is squared (\( x^2 \)). In solving mathematical problems, transforming a complicated expression into a quadratic form can simplify the solving process.

In the exercise we examined, we derived the quadratic equation \( x^2 - 9x + 14 = 0 \) after eliminating the square roots from the original expression. Solving this type of equation usually involves either:

• Factoring, if it's factorable.
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Completing the square.

Through factoring, we identified the factors \((x - 7)(x - 2) = 0\), leading us to potential solutions: \( x = 7 \) and \( x = 2 \). These solve the quadratic equation, but we further verify them in the context of their applicability to the original problem to ensure they are real solutions.

Square root equations

Equations that contain square roots are known as square root equations. Solving them involves careful techniques to systematically remove said roots. The key challenge is ensuring the process doesn't introduce extraneous solutions—solutions that arise from algebraic manipulation but don't satisfy the original equation.

In our example problem, we encountered the equation \( \sqrt{x+\sqrt{x+2}} = 2 \). This involved working through multiple nested square roots. To eradicate these roots, we:

• Square both sides of the equation to eliminate the outer root, yielding \( x + \sqrt{x+2} = 4 \).
• Isolate the remaining square root, followed by another squaring step.
• Simplify the resulting equation to make it solvable, in this case, converting to a quadratic equation.

After getting potential solutions, verify them by plugging them back into the original equation. This avoids accepting invalid results introduced during squaring. In this case, only \( x = 2 \) was valid, as when plugged back in it satisfied the initial square root equation.