Problem 44 - Algebra 1

Question

Solve the equation. Check for extraneous solutions. $$2 x=\sqrt{-13 x-10}$$

Step-by-Step Solution

Answer

The equation \(2x = \sqrt{-13x - 10}\) has no valid solutions since both potential solutions \(-\frac{5}{4}\) and \(-2\) are extraneous.

Step 1: Squaring both sides of the equation

To eliminate the square root, square both sides of the equation to get a quadratic equation. So \((2x)^2 = (\sqrt{-13x-10})^2\) which simplifies to \(4x^2 = -13x - 10\).

Step 2: Solving the quadratic equation

Rearrange terms and set the equation to zero: \(4x^2 + 13x + 10 = 0\). Factoring this quadratic equation yields \((4x+5)(x+2) = 0\). Setting each factor equal to zero gives two potential solutions: \(x=-\frac{5}{4}\) and \(x=-2\)

Step 3: Checking for extraneous solutions

Substitute these potential solutions back into the original equation to verify if they are valid. Substituting \(x=-\frac{5}{4}\) comes out false as \(-\frac{5}{2} \neq \sqrt{-13(-\frac{5}{4})-10}\). Substituting \(x=-2\) gives us \(-4=\sqrt{-13(-2)-10}\) which is also false. After squaring both sides, it comes out to be 16 equals to 36, which is false. Therefore, both solutions are extraneous, and the original equation has no solution.

Key Concepts

Understanding Extraneous Solutions

An extraneous solution is a solution derived from an algebraic manipulation that doesn't satisfy the original equation. This often occurs when working with equations involving square roots or rational expressions.
When both sides of an equation are squared, new solutions may be introduced that don't satisfy the starting equation, leading to these extraneous results.
In our problem, after factoring and solving the quadratic equation, we had two potential solutions: \(x = -\frac{5}{4}\) and \(x = -2\). However, substituting these values back into the original equation revealed neither worked, marking them as extraneous.

• Always plug solutions back into the original equation to verify their validity.
• Look out for signs of extraneous solutions in problems involving square roots or fractions.

Factoring Quadratic Equations

Factoring quadratics is a common method to solve quadratic equations of the form \(ax^2 + bx + c = 0\). Here are some steps to factor quadratics effectively.
In our exercise, we rearranged \(4x^2 + 13x + 10 = 0\) and factored it to get \((4x + 5)(x + 2) = 0\). These factors were identified by looking for two numbers that multiply to \(4 \times 10 = 40\) and add to 13.

• This method is efficient when the quadratic is factorable.
• Always check your factors by expanding them back to ensure accuracy.

If direct factoring is hard, you can use the quadratic formula as an alternative. Remember, factoring is not the end; always proceed to verify your findings.

Checking Solutions

Checking solutions is a crucial step in solving equations to confirm the correctness of your results. It's especially important when dealing with manipulations like squaring both sides of an equation.
In this task, verifying the solutions by substituting them back into the original equation showed us that \(x = -\frac{5}{4}\) and \(x = -2\) were not valid, despite being potential solutions derived from the quadratic equation.
To check effectively:

• Always substitute back into the original equation, not the transformed one.
• If the equation doesn't balance, the solution is extraneous.

This process ensures that any extraneous solutions are identified and removed, aligning your final answer with the problem's true solution set.