Problem 100 - Intermediate Algebra

Question

Solve. \(x^{2}+6 x=4 \sqrt{x^{2}+6 x}\)

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 2\), and \(x = -8\).

Step 1: Identify Variables and Simplify

Firstly, identify the equation and variables. The given equation is \(x^{2} + 6x = 4\sqrt{x^{2} + 6x}\). Let's set \(y = x^2 + 6x\) to simplify the equation. Hence, it becomes \(y = 4\sqrt{y}\).

Step 2: Solve for \(y\)

To solve \(y = 4\sqrt{y}\), square both sides to eliminate the square root: \(y^2 = 16y\). Rearrange this to form a quadratic equation: \(y^2 - 16y = 0\).

Step 3: Factor the Quadratic Equation

The equation \(y^2 - 16y = 0\) can be factored as \(y(y - 16) = 0\). Therefore, the solutions for \(y\) are \(y = 0\) or \(y = 16\).

Step 4: Substitute Back to Solve for \(x\)

Recall \(y = x^2 + 6x\), substitute \(y = 0\) and \(y = 16\) back to find \(x\). For \(y = 0\), \(x^2 + 6x = 0\) factors to \(x(x + 6)=0\), giving solutions \(x = 0\) and \(x = -6\). For \(y = 16\), solve \(x^2 + 6x - 16 = 0\) using the quadratic formula.

Step 5: Use Quadratic Formula for \(x^2 + 6x - 16 = 0\)

The quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) will be used. Here, \(a = 1\), \(b = 6\), and \(c = -16\). Compute the discriminant: \(b^2 - 4ac = 36 + 64 = 100\). Therefore, \(x = \frac{-6 \pm 10}{2}\), giving \(x = 2\) and \(x = -8\).

Step 6: Verify the Solutions

Verify each potential solution: \(x = 0\), \(x = -6\), \(x = 2\), and \(x = -8\) in the original equation. Substituting provides valid solutions: \(x = 0\), \(x = 2\), and \(x = -8\). \(x = -6\) doesn't satisfy the equation, as it causes an invalid expression under the square root.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a crucial skill in solving equations. It involves rearranging and transforming equations to isolate variables and simplify the expressions.
In our exercise, we start with the equation:

• \(x^2 + 6x = 4\sqrt{x^2 + 6x}\)

To simplify, let's introduce a substitution:

• Let \(y = x^2 + 6x\).

This gives us a new equation in terms of \(y\):

• \(y = 4\sqrt{y}\).

By squaring both sides, we eliminate the square root and simplify our equation to a more familiar quadratic form:

• \(y^2 = 16y\).
• Rewriting gives \(y^2 - 16y = 0\).

This process highlights the power of substitution and squaring techniques to ease complex algebraic manipulations.

Quadratic Formula

The quadratic formula is a dependable tool for solving quadratic equations when factoring is difficult or infeasible. It provides a universal method to find the roots of any equation of the form \(ax^2 + bx + c = 0\).
The formula is:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

In our exercise, we encounter the equation resulting from substituting back the solution for \(y = 16\):

• \(x^2 + 6x - 16 = 0\).

By applying the quadratic formula:

• Here, \(a = 1\), \(b = 6\), \(c = -16\).
• Compute the discriminant: \(b^2 - 4ac = 36 + 64 = 100\).
• Hence, \(x = \frac{-6 \pm 10}{2}\).

These calculations yield the solutions \(x = 2\) and \(x = -8\). The quadratic formula streamlines the finding of solutions, ensuring precision even with complex numbers.

Factoring

Factoring is a method of breaking down complex expressions into simpler factors that can be solved individually. It allows us to handle quadratic equations efficiently when the factors are straightforward.
In our exercise, after using substitution, we deal with the equation:

• \(y^2 - 16y = 0\).

This expression can be factored by taking out the common factor of \(y\):

• \(y(y - 16) = 0\).

This gives us the simple solutions:

• \(y = 0\) or \(y = 16\).

Factoring allows us to quickly identify the roots of the quadratic by observing when each factor equals zero.

Solution Verification

Verifying solutions ensures that the values obtained are correct and suitable for the original problem setup.
In our problem, once we have the potential solutions for \(x\):

• \(x = 0, -6, 2, -8\).

We must substitute each back into the original equation:

• \(x^2 + 6x = 4\sqrt{x^2 + 6x}\).

Checking each value:

• \(x = 0\): Satisfies as both sides zero out.
• \(x = 2\): Validates the equation without contradictions.
• \(x = -8\): Also fits correctly upon substitution.
• \(x = -6\): Leads to a square root of a negative, which is not permissible (unless using complex numbers).

Thus, the verified solutions are \(x = 0, 2, -8\), demonstrating the importance of this step to ensure all conditions of the problem are met.