Problem 64 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\sqrt[4]{x^{2}+6 x}=2$$

Step-by-Step Solution

Answer

The solutions are x = 2 and x = -8.

Step 1: Understand the Given Equation

The given equation is \[\sqrt[4]{x^{2}+6x} = 2\]. Recognize that the left side of the equation involves a fourth root.

Step 2: Eliminate the Fourth Root

To remove the fourth root, raise both sides of the equation to the power of 4: \[\bigg(\sqrt[4]{x^{2}+6x}\bigg)^4 = 2^4\]. This simplifies to \[\big(x^{2}+6x\big) = 16\].

Step 3: Form a Quadratic Equation

After simplifying, the equation becomes \[\big(x^{2}+6x - 16 = 0\]. This is a standard quadratic equation in the form \[ax^2 + bx + c = 0\].

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \[\big(x^{2}+6x - 16 = 0\]. Use the quadratic formula \[\big(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\big)\] where \[a = 1, b = 6, c = -16\].

Step 5: Apply the Quadratic Formula

Substitute the coefficients into the formula: \[\big(x = \frac{-6 \pm \sqrt{6^2-4 \cdot 1 \cdot (-16)}}{2 \cdot 1}\big)\], which simplifies to \[x = \frac{-6 \pm \sqrt{36+64}}{2}\big)\], \[\big(x = \frac{-6 \pm \sqrt{100}}{2}\big)\],

Step 6: Simplify the Quadratic Solution

This further simplifies to \[x = \frac{-6 \pm 10}{2}\]. Therefore, we have two solutions: \[x = \frac{-6 + 10}{2} = 2\] and \[x = \frac{-6 - 10}{2} = -8\].

Step 7: Verify the Solutions

Substitute the solutions back into the original equation to verify: For \[x = 2, \sqrt[4]{2^2 + 6 \cdot 2} = \sqrt[4]{4 + 12} = \sqrt[4]{16} = 2\], so \[x = 2\] is correct. For \[x = -8, \sqrt[4]{(-8)^2 + 6 \cdot (-8)} = \sqrt[4]{64 - 48} = \sqrt[4]{16} = 2\], so \[x = -8\] is also correct.

Key Concepts

Fourth Root

The fourth root of a number is an important concept in algebra. It refers to a value that, when raised to the fourth power, gives the original number. For example, the fourth root of 16 is 2, because \(2^4=16\). When solving the given equation, \(\root[4]{x^2 + 6x} = 2\), our goal is to isolate the variable inside the fourth root.

To eliminate the fourth root, raise both sides of the equation to the power of 4. This operation effectively cancels out the root, leading us to a much simpler equation:
\(\big(\root[4]{x^2 + 6x}\big)^4 = 2^4\).

Thus, the equation simplifies to:
\(x^2 + 6x = 16\). This allows us to move on to the next steps where solving becomes easier.

Quadratic Formula

The quadratic formula is commonly used to solve quadratic equations of the form \ax^2 + bx + c = 0\. It is expressed as:
\(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

In our exercise, we simplified the given equation to \(x^2 + 6x - 16 = 0\), a standard quadratic form where \(a = 1, b = 6, c = -16\).

Next, substitute the values into the quadratic formula:
\(x = \frac{-6 \pm \sqrt{6^2-4*1*(-16)}}{2*1}\).

This simplifies to:
\(x = \frac{-6 \pm \sqrt{36 + 64}}{2}\), which further simplifies to:
\(x = \frac{-6 \pm 10}{2}\).

Hence, we find two possible solutions for \(x\):

• \(x = \frac{-6 + 10}{2} = 2\)
• \(x = \frac{-6 - 10}{2} = -8\)

Verifying Solutions

Verifying solutions is a crucial step to ensure the accuracy of the solved values. We need to substitute the solutions back into the original equation, \(\root[4]{x^2 + 6x} = 2\), to check their validity.

First, for \(x = 2\):
\(\root[4]{2^2 + 6*2} = \root[4]{4 + 12} = \root[4]{16} = 2\).
This confirms that \(x = 2\) is indeed a correct solution.

Next, for \(x = -8\):
\(\root[4]{(-8)^2 + 6*(-8)} = \root[4]{64 - 48} = \root[4]{16} = 2\).
This confirms that \(x = -8\) is also a correct solution.

Therefore, both solutions \(x = 2\) and \(x = -8\) satisfy the original equation, proving their correctness.