Problem 27 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Square Both Sides

To eliminate the square root, we square both sides of the equation. Therefore, we have: \((\sqrt{x+2})^2 = x^2\). This simplifies to \(x+2 = x^2\).

Step 2: Rearrange Equation

Rearrange the equation to bring all terms on one side and set it to zero. This gives us: \(x^2 - x - 2 = 0\).

Step 3: Factor the Quadratic Equation

Factor the quadratic equation \(x^2 - x - 2\). Find two numbers that multiply to -2 and add to -1. These numbers are -2 and 1. Thus, the equation factors to \((x-2)(x+1) = 0\).

Step 4: Solve Each Factor

Set each factor equal to zero and solve for \(x\). \(x - 2 = 0\) gives \(x = 2\) and \(x + 1 = 0\) gives \(x = -1\).

Step 5: Verify Solutions

Substitute the potential solutions back into the original equation \(\sqrt{x+2} = x\) to verify. For \(x = 2\), substitute to get \(\sqrt{2+2} = 2\), which is true. For \(x = -1\), substitute to get \(\sqrt{-1+2} = -1\), which simplifies to \(\sqrt{1} = -1\), which is false. So, \(x = -1\) is not a valid solution.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest exponent of the variable is two. Generally, these equations are in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The term "quadratic" comes from "quad," meaning square, as the variable term is squared.

To solve a quadratic equation, you can use several methods such as:

• Factoring: Express the quadratic as a product of two binomials.
• Quadratic formula: Use the formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) if factoring is challenging or impossible.
• Completing the square: Rearrange the equation into a perfect square trinomial.

For this exercise, we successfully factored the equation \(x^2 - x - 2 = 0\) into \((x-2)(x+1) = 0\). Factoring makes it easy to find potential solutions by setting each factor to zero:

\((x-2) = 0\) and \((x+1) = 0\).

The quadratic equation principle is an essential arithmetic foundation for solving polynomial equations and understanding more complex algebraic concepts.

Working with Square Roots

A square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\). For example, the square root of 9 is 3, as \(3 \times 3 = 9\). The square root is represented with the symbol \(\sqrt{ }\).

In this equation, we started with \(\sqrt{x+2} = x\). To remove the square root and make solving the equation easier, we squared both sides: \((\sqrt{x+2})^2 = x^2\). Squaring gets rid of the square root and reveals a quadratic equation \(x+2 = x^2\).

However, squaring can lead to extraneous solutions, which are apparent solutions that don’t actually satisfy the original equation. That's why verifying solutions is crucial. In this exercise, squaring introduced \(x = -1\) as a potential solution, which didn't satisfy the original equation upon verification: \(\sqrt{-1+2} eq -1\).

When dealing with square roots in equations, always remember to check your solutions to ensure they truly work in the original context.

Finding Real Solutions

Real solutions, in terms of equations, are the solutions that are actual numbers you can locate on a number line, as opposed to imaginary or complex solutions. In this problem, we aimed to find real solutions that satisfy \(\sqrt{x+2} = x\).

When we squared both sides and rearranged the terms into a quadratic equation, we obtained potential solutions \(x = 2\) and \(x = -1\) from \((x-2)(x+1) = 0\).

To ensure we only include valid real solutions, it's necessary to substitute back into the original equation. Here,

• \(x = 2\) yielded \(\sqrt{4} = 2\), which is true,
• whereas \(x = -1\) produced \(\sqrt{1} eq -1\), making it invalid in this context.

Proper verification helps exclude extraneous solutions and confirms the authenticity of your real solutions.

This clarification process emphasizes why validation is a critical final step in solving equations, especially when dealing with transformations like squaring that can expand the set of solutions beyond those that are genuine.