Problem 39 - Algebra A Combined Function

Question

Solve. $$ x-\sqrt{1-x}=-5 $$

Step-by-Step Solution

Answer

The solution is \( x = -3 \).

Step 1: Isolate the Radical

To solve the equation \( x - \sqrt{1-x} = -5 \), the first step is to isolate the radical term. We do this by adding \( \sqrt{1-x} \) to both sides of the equation:\[ x = -5 + \sqrt{1-x} \]

Step 2: Rearrange Equation

Our goal is to eliminate the square root, so we rearrange the equation to isolate the square root term on one side:\[ \sqrt{1-x} = x + 5 \]

Step 3: Eliminate the Radical

Square both sides to get rid of the square root. Remember, both sides of the equation need to be squared:\[ (\sqrt{1-x})^2 = (x + 5)^2 \]This simplifies to:\[ 1 - x = (x + 5)^2 \]

Step 4: Expand and Simplify

Expand the right side of the equation:\[ 1 - x = x^2 + 10x + 25 \]Reorganize to form a quadratic equation:\[ 0 = x^2 + 10x + 25 - 1 + x \]Simplify:\[ 0 = x^2 + 11x + 24 \].

Step 5: Solve the Quadratic Equation

Now, solve the quadratic equation \( x^2 + 11x + 24 = 0 \). Factor the equation:\[ (x + 3)(x + 8) = 0 \]This gives us the solutions:\[ x + 3 = 0 \; \Rightarrow \; x = -3 \]\[ x + 8 = 0 \; \Rightarrow \; x = -8 \]

Step 6: Check Potential Solutions

Substitute both values back into the original equation to check for extraneous solutions:1. For \( x = -3 \):\[-3 - \sqrt{1 - (-3)} = -3 - \sqrt{4} = -3 - 2 = -5 \]This checks out as valid.2. For \( x = -8 \):\[-8 - \sqrt{1 - (-8)} = -8 - \sqrt{9} = -8 - 3 = -11 eq -5 \]This is not a valid solution.

Key Concepts

Isolating the Radical

Isolating the radical term is a crucial step in solving radical equations like \(x - \sqrt{1-x} = -5\). The process involves getting the radical alone on one side of the equation.

• Starting Point: Begin with the original equation, \(x - \sqrt{1-x} = -5\).
• Adjust the Equation: Add the square root term from the left side to the right side: \(x = -5 + \sqrt{1-x}\).
• Check Your Work: Confirm that the radical is truly isolated on one side.

Once the radical is isolated, the next steps become clearer, setting the stage for successfully eliminating the square root.

Quadratic Equation Solving

With the radical isolated, the next logical step is to eliminate it by squaring both sides of the equation. This transforms your problem into a quadratic equation, which can then be tackled using traditional algebraic methods.

• Square Both Sides: Eliminate the square root by squaring the entire equation: \((\sqrt{1-x})^2 = (x + 5)^2\).
• Simplify: After expansion, we have \(1 - x = x^2 + 10x + 25\).
• Formulate Quadratic Equation: Rearrange it to form a familiar quadratic equation: \(0 = x^2 + 11x + 24\).

Understanding and applying these steps ensures you handle the quadratic equation methodically, paving the way for solving it using factoring or other methods.

Extraneous Solutions

In radical equations, sometimes squaring both sides can introduce solutions that don't actually satisfy the original equation. These are called extraneous solutions.

• Identify Extraneous Solutions: Substitute each obtained solution back into the original equation.
• Verification: For \(x = -3\), plugging back into the original equation shows it holds true (\(-3 - \sqrt{4} = -5\)).
• Discard Unwanted Solutions: For \(x = -8\), substitution does not solve the original equation (resulting in \(-11\), not \(-5\)).

This process is essential to ensure that all solutions are valid and relevant, supporting a complete and accurate problem-solving journey.

Factoring Quadratics

Factoring is a straightforward technique that helps in finding the roots of quadratic equations, and it’s often simpler than using the quadratic formula.

• Identify the Quadratic: First, ensure your equation is set to zero: \(x^2 + 11x + 24 = 0\).
• Factor the Terms: Look for two numbers that add up to 11 and multiply to 24, resulting in \((x + 3)(x + 8) = 0\).
• Solve for Each Root: Set each factor equal to 0 to find the solutions: \(x = -3\) and \(x = -8\).

Factoring quadratics simplifies the solution process and directly addresses the roots, laying a clear path toward determining the valid solutions.