Problem 31 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \sqrt{5-x}+1=x-2 $$

Step-by-Step Solution

Answer

The real solution to the equation is \(x = 4\).

Step 1: Isolate the Square Root

Let's first isolate the square root term on one side of the equation. Start with:\[\sqrt{5-x} + 1 = x - 2\]Subtract 1 from both sides to get:\[\sqrt{5-x} = x - 3\]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation:\[(\sqrt{5-x})^2 = (x-3)^2\]This simplifies to:\[5-x = (x-3)^2\]

Step 3: Expand the Right Side

Now expand \((x-3)^2\):\[5-x = x^2 - 6x + 9\]

Step 4: Rearrange into a Quadratic Equation

Rearrange the terms to form a standard quadratic equation:\[x^2 - 5x + 4 = 0\]

Step 5: Solve the Quadratic Equation

To solve the quadratic equation \(x^2 - 5x + 4 = 0\), we can factor it. The factors of 4 that sum to 5 are 1 and 4, so:\[(x-4)(x-1) = 0\]Set each factor to zero:\[x-4 = 0 \quad \text{or} \quad x-1 = 0\]Thus, \(x = 4\) or \(x = 1\).

Step 6: Verify the Solutions

Substitute \(x = 1\) and \(x = 4\) back into the original equation to check their validity:For \(x = 1\):\[\sqrt{5-1} + 1 = 1 - 2\]\[2 + 1 = -1\] which is false.For \(x = 4\):\[\sqrt{5-4} + 1 = 4 - 2\]\[1 + 1 = 2\] which is true.Therefore, only \(x = 4\) is a solution.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree in the form \( ax^2 + bx + c = 0 \). The coefficients \( a \), \( b \), and \( c \) are real numbers, and \( a eq 0 \). In the given problem, through a sequence of transformations—including isolating the square root and expanding the squared terms—the equation \( \sqrt{5-x} + 1 = x - 2 \) was rearranged into a quadratic equation: \( x^2 - 5x + 4 = 0 \).

This form is crucial because it allows us to utilize various methods for finding solutions, such as factoring, completing the square, or using the quadratic formula. In this case, the quadratic \( x^2 - 5x + 4 \) was solved by factoring. Factoring is feasible when the quadratic is easy to decompose into two binomial expressions, like \((x-4)(x-1) = 0\).

By setting each factor to zero, we find two potential solutions: \( x = 4 \) and \( x = 1 \). Every solution, however, should be verified by substituting back into the original equation to ensure it holds true, as seen in later steps of our process.

Square Root Isolation

Square root isolation is a strategic step undertaken to simplify the equation by having the square root on one side. In our exercise, the goal was to isolate the \( \sqrt{5-x} \) term to proceed with solving the equation. By starting with:\[ \sqrt{5-x} + 1 = x - 2 \]and subtracting 1 from both sides, the equation transforms to:\[ \sqrt{5-x} = x - 3 \].

Once isolated, the next logical step is to eliminate the square root by squaring both sides. This step is critical because it converts a square root equation into a polynomial equation, specifically a quadratic one in this context. Both sides are squared, resulting in:\[( \sqrt{5-x} )^2 = (x-3)^2\]which simplifies to:\[5-x = x^2 - 6x + 9\].

Isolating the square root is essential whenever we encounter an equation with a square root term. This step not only simplifies the equation but is also imperative for making the equation solvable through traditional algebraic methods.

Solution Verification

Solution verification ensures that the potential solutions satisfy the original equation. Due to the nature of squaring both sides during the process, extraneous solutions can arise. For this exercise, both solutions derived from solving the quadratic equation—namely \( x = 4 \) and \( x = 1 \)—must be checked within the initial square root equation to verify their validity.

Substituting \( x = 1 \) back into the original equation:\[ \sqrt{5-1} + 1 = 1 - 2 \]results in:\[ 2 + 1 = -1 \].

The left side doesn't equal the right, revealing \( x = 1 \) as an extraneous solution. Conversely, substituting \( x = 4 \):\[ \sqrt{5-4} + 1 = 4 - 2 \]results in:\[ 1 + 1 = 2 \].

This holds true, verifying that \( x = 4 \) is indeed a valid solution. Verification is crucial in mathematical problem-solving as it affirms that the values obtained are true solutions of the given equation.