Problem 37 - Algebra 1

Question

Solve the equation. Check for extraneous solutions. $$x=\sqrt{-4 x-4}$$

Step-by-Step Solution

Answer

The given equation has no solutions as \(x = -2\) is an extraneous solution.

Step 1: Isolate the square root

The square root function can be isolated on one side by squaring both sides of the equation. Squaring the left side \(x\) gives \(x^2\), and squaring the square root on the right side will remove the square root, leaving us with \(-4x - 4\). Our new equation will be \(x^2 = -4x -4\)

Step 2: Convert to quadratic equation

The equation \(x^2 = -4x - 4\) can be made into a standard quadratic equation by adding \(4x + 4\) to both sides of the equation. This step yields \(x^2 + 4x + 4 = 0\)

Step 3: Factor the quadratic equation

The quadratic equation \(x^2 + 4x + 4 = 0\) can be factored as \((x + 2)^2 = 0\) or it could be solved by applying quadratic formula. However, factoring the quadratic equation is simpler in this case. After factoring, solve for \(x\) by setting \(x + 2 = 0\), which yields \(x = -2\)

Step 4: Check for extraneous solutions

Substitute \(x = -2\) into the original equation to check if it is an extraneous solution. Substituting into the original equation yields \(-2 = \sqrt{-4*-2 - 4}\), which simplifies to \(-2 = \sqrt{4}\). This equation does not hold, meaning that \(x = -2\) is an extraneous solution and thus not a solution to the equation.

Key Concepts

Understanding Extraneous Solutions

When solving equations involving square roots, you may encounter extraneous solutions. These are solutions that arise from the algebraic manipulation of the equation but do not actually satisfy the original equation.

In this exercise, when we squared both sides of the equation \(x = \sqrt{-4x - 4}\), we introduced the possibility of extraneous solutions. Always substitute back into the original equation to verify your solutions:

• For example, if we find \(x = -2\), substituting back, we find that \(-2 = \sqrt{4}\) is not true.
• This means \(x = -2\) is not a valid solution but rather an extraneous one.

Checking for extraneous solutions is a crucial step in solving radical equations. Never skip it!

Quadratic Equations Explained

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They appear frequently in algebra.

In our problem, we transformed \(x = \sqrt{-4x - 4}\) into a quadratic equation \(x^2 + 4x + 4 = 0\) by squaring both sides and rearranging. Solving this quadratic equation can be done by:

• Factoring the quadratic expression if possible.
• Using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Quadratics are versatile and can model many real-world phenomena. Mastering them helps in understanding more complex mathematical concepts.

Factoring Quadratics with Ease

Factoring is a technique for solving quadratic equations by expressing the quadratic expression as a product of linear factors.

For the equation \(x^2 + 4x + 4 = 0\), we noticed that it can be factored as \((x + 2)(x + 2) = 0\), or \((x + 2)^2 = 0\).

• Set each factor equal to zero: \(x + 2 = 0\).
• Solve for \(x\) to find the roots.

Factoring simplifies the problem and is usually quicker than the quadratic formula when applicable. Practice recognizing patterns like perfect squares to become efficient at factoring quadratics.