Problem 15 - Algebra A Combined Function

Question

Solve. $$ \sqrt{13-x}=x-1 $$

Step-by-Step Solution

Answer

x = 4 is the solution.

Step 1: Isolate the Square Root

The equation given is \( \sqrt{13-x} = x - 1 \). Notice that the square root is on one side of the equation by itself. Our initial condition is satisfied, so no isolation is needed for the square root term.

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root. This gives us:\[ (\sqrt{13-x})^2 = (x-1)^2 \]This simplifies to:\[ 13-x = x^2 - 2x + 1 \]

Step 3: Move All Terms to One Side

Rearrange the equation to form a quadratic equation by moving all terms to one side:\[ x^2 - 2x + 1 + x - 13 = 0 \]Simplify this to:\[ x^2 - x - 12 = 0 \]

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( x^2 - x - 12 = 0 \) by finding two numbers that multiply to \(-12\) and add to \(-1\):\[ x^2 - 4x + 3x - 12 = 0 \]which can be factored as:\[ (x - 4)(x + 3) = 0 \]

Step 5: Solve the Factors

Solve for \( x \) from the factored form:1. \( x - 4 = 0 \) gives \( x = 4 \).2. \( x + 3 = 0 \) gives \( x = -3 \).

Step 6: Check for Extraneous Solutions

Substitute the solutions back into the original equation to check for extraneous solutions:- For \( x = 4 \): \[ \sqrt{13-4} = 4 - 1 \rightarrow 3 = 3 \text{ (valid)} \]- For \( x = -3 \): \[ \sqrt{13 - (-3)} = -3 - 1 \rightarrow \sqrt{16} eq -4 \] (invalid, as the square root is always non-negative).Only \( x = 4 \) is a valid solution.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental aspect of algebra that you'll frequently encounter. These are polynomial equations of the second degree, typically written in the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The equation has:

• Two real, equal or complex solutions, also known as roots.
• A signature 'U' shaped curve called a parabola when graphed.

In our exercise, this equation materializes after squaring both sides to eliminate the square root, forming \( x^2 - x - 12 = 0 \). Notice how the quadratic nature emerges from the rearrangement of terms. Solving quadratics usually involves factoring, completing the square, or using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Each method has its use cases depending on the specific equation and the ease of calculation.

Identifying and Understanding Extraneous Solutions

Extraneous solutions can appear when performing operations like squaring both sides of an equation, as it can introduce solutions that don't satisfy the original equation. They are not actual solutions to the problem but rather by-products of the manipulation.

• These solutions may arise because certain mathematical operations are not reversible.
• They must be checked against the original equation to confirm validity.

For instance, in the given problem, squaring the equation \( \sqrt{13-x} = x-1 \) and solving led us to two potential solutions: \( x = 4 \) and \( x = -3 \). Checking these, only \( x = 4 \) is correct as substituting \( x = -3 \) into the original equation gives a contradiction, thus it's extraneous. Always verify solutions in radical equations to filter out any extraneous ones.

Factoring as a Solving Strategy

Factoring is a strategic approach when solving quadratic equations where the equation is set to zero. This involves rewriting the quadratic expression as a product of two binomials.

• Find two numbers that multiply to \( ac \) (the product of the quadratic and constant terms) and add to \( b \) (the coefficient of \( x \)).
• Rewrite the middle term using these two numbers and group terms to factor by grouping.
• Solve the resulting linear equations.

In our example, for \( x^2 - x - 12 = 0 \), the goal was to find factors of \(-12\) that add up to \(-1\). The integers \(-4\) and \(+3\) fit, allowing us to express it as \( (x-4)(x+3) = 0 \). Each factor equates to zero providing potential solutions. Factoring is efficient for simple quadratics and helps visualize roots, making it a favored technique before considering more complex methods.