Problem 91 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{-5 x+24}=6-x $$

Step-by-Step Solution

Answer

The solutions are \(x = 3\) and \(x = 4\), with no extraneous solutions.

Step 1: Eliminate the Square Root

To remove the square root, square both sides of the equation:\( (\sqrt{-5x + 24})^2 = (6-x)^2 \). This results in the equation: \(-5x + 24 = (6-x)^2\).

Step 2: Expand the Squared Term

Expand the right side using the formula \((a-b)^2 = a^2 - 2ab + b^2\). Thus, \((6-x)^2 = 36 - 12x + x^2\). The equation becomes \(-5x + 24 = 36 - 12x + x^2\).

Step 3: Rearrange the Equation

Rearrange all terms to one side to form a quadratic equation: \(x^2 - 7x + 12 = 0\).

Step 4: Solve the Quadratic Equation

Factor the quadratic equation. Look for two numbers that multiply to 12 and add to -7: (-3) and (-4). Therefore, the factored equation is \((x-3)(x-4) = 0\). This gives potential solutions \(x = 3\) and \(x = 4\).

Step 5: Check for Extraneous Solutions

Substitute \(x = 3\) back into the original equation: \(\sqrt{-5(3) + 24}=6-3\). This simplifies to \(\sqrt{9} = 3\), which is correct. Substitute \(x=4\) back: \(\sqrt{-5(4)+24}=6-4\) simplifies to \(\sqrt{4}=2\), which is also correct. Both solutions are valid.

Key Concepts

Extraneous Solutions

When solving equations, especially those involving square roots, extraneous solutions can sometimes appear. These are solutions that emerge from the process of solving the problem, but do not satisfy the original equation. This occurs because some operations, like squaring both sides, can introduce solutions that aren't actually valid.

For instance, consider the equation \( \sqrt{-5x + 24} = 6 - x \). Once both sides are squared, we get a quadratic equation that might suggest multiple solutions. However, these need to be verified against the original equation.

To do this, plug each proposed solution back into the original equation:

• If both sides satisfy the equation, it is a true solution.
• If they do not, this indicates an extraneous solution.

Checking for extraneous solutions is a crucial last step to ensure the validity of your results.

Factoring Quadratics

Factoring quadratic equations is a method of solving these equations by expressing them as a product of linear factors. This is done by finding two numbers that multiply to give the constant term and add up to give the coefficient of the linear term.

Consider the quadratic equation in the step-by-step solution: \( x^2 - 7x + 12 = 0 \). We need to find two numbers that multiply to 12 and add to -7.

• The numbers are -3 and -4.
• Thus, the equation can be factored as \( (x - 3)(x - 4) = 0 \).

By solving each factor set to zero, \( x - 3 = 0 \) and \( x - 4 = 0 \), we find the potential solutions: \( x = 3 \) and \( x = 4 \).
Factoring is a powerful tool when the quadratic equation can be easily factored, combining simplicity with effectiveness to provide solutions quickly.

Solving Square Root Equations

Square root equations, such as \( \sqrt{-5x + 24} = 6 - x \), require careful handling to remove the radical and solve the equation. A common strategy is to eliminate the square root by squaring both sides of the equation. This transforms it into a polynomial equation that is typically easier to solve.

However, this process can introduce complications such as extraneous solutions. After squaring, solve the resulting polynomial by using methods such as factoring.

Remember to:

• Isolate the square root on one side before squaring.
• Squaring every term can change the root of the equation, sometimes giving false results.
• Always check your potential solutions against the original equation to ensure accuracy.

This careful approach helps to properly manage the unique challenges posed by square root equations and prevents errors that could arise from incorrect solutions.