Problem 22 - Intermediate Algebra

Question

Solve each radical equation. Don't forget, you must check potential solutions. $$ 3 \sqrt{2 x}=x+4 $$

Step-by-Step Solution

Answer

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

Step 1: Isolate the Radical Expression

To solve the given radical equation, first ensure the radical is isolated on one side of the equation. The given equation is already in this form: \[3 \sqrt{2x} = x + 4\]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation. This transforms the equation as follows:\[(3 \sqrt{2x})^2 = (x + 4)^2\]Simplifying both sides, you get:\[9 \cdot 2x = x^2 + 8x + 16\]\[18x = x^2 + 8x + 16\]

Step 3: Rearrange the Equation

Move all terms to one side of the equation to form a quadratic equation:\[0 = x^2 + 8x + 16 - 18x\]Simplifying gives:\[0 = x^2 - 10x + 16\]

Step 4: Solve the Quadratic Equation

Use the quadratic formula to solve for \(x\). The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 1\), \(b = -10\), and \(c = 16\). Substituting these values gives:\[x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1}\]\[x = \frac{10 \pm \sqrt{100 - 64}}{2}\]\[x = \frac{10 \pm \sqrt{36}}{2}\]\[x = \frac{10 \pm 6}{2}\]So, \(x = 8\) or \(x = 2\).

Step 5: Check Potential Solutions

Substitute \(x = 8\) and \(x = 2\) back into the original equation to check for potential extraneous solutions.For \(x = 8\):\[3 \sqrt{2 \cdot 8} = 8 + 4 \]\[3 \times 4 = 12\]This is correct.For \(x = 2\):\[3 \sqrt{2 \cdot 2} = 2 + 4 \]\[3 \times 2 = 6\]This is also correct.

Key Concepts

Quadratic Equation

A quadratic equation is an equation that can be written in the standard form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). These equations are known for their distinctive parabola-shaped graph. In our exercise, after isolating the radical and manipulating the original radical equation through squaring both sides, we obtain the quadratic equation:
\[x^2 - 10x + 16 = 0\]
This transformation occurs when linear equations or expressions with roots are squared, which can introduce additional solutions that are not valid due to the nature of the quadratic form.

• When all terms are on one side, like in the form \(x^2 - 10x + 16 = 0\), we can visually or algebraically work with it using various methods like completing the square or, as we used here, the quadratic formula.

It is important to note that while solving these equations might be straightforward once it’s in the quadratic form, one must always be wary of solutions that might not satisfy the original equation.

Extraneous Solutions

Extraneous solutions arise frequently when solving radical equations or any equation that involves manipulation such as squaring both sides. These are solutions that emerge from the algebraic process of solving the equation but do not satisfy the original equation upon substitution.
In the problem at hand, after solving the quadratic equation, we found potential solutions \(x = 8\) and \(x = 2\). However, verifying these solutions against the original equation \(3 \sqrt{2x} = x + 4\) is crucial. Each solution must be substituted back into this original equation:

• For \(x = 8\): The substitution holds true since both sides equal 12.
• For \(x = 2\): The substitution also works out since both sides equal 6.

The process of checking these solutions ensures that no extraneous solutions have been accepted. This step is critical whenever transformations or additional operations have been applied to an equation.

Quadratic Formula

The quadratic formula is a powerful tool that allows us to find the roots of any quadratic equation \(ax^2 + bx + c = 0\). It is expressed as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In our example, after rewriting the equation into its standard quadratic form \(x^2 - 10x + 16 = 0\), the quadratic formula was employed:

• Identified coefficients: \(a = 1\), \(b = -10\), \(c = 16\).
• Calculated the discriminant: \((-10)^2 - 4 \times 1 \times 16 = 36\).
• Inserted these into the formula, yielding potential solutions \(x = 8\) or \(x = 2\).

The discriminant, \(b^2 - 4ac\), gives insight into the nature of the solutions. A positive value, such as in this exercise, indicates two real solutions, while a discriminant of zero would mean just one real solution.