Problem 29 - College Algebra Essentials

Question

Use the quotient rule to simplify the expressions in Exercises \(23-32\) Assume that \(x>0\) $$\frac{\sqrt{150 x^{4}}}{\sqrt{3 x}}$$

Step-by-Step Solution

Answer

\(\frac{5\sqrt{6} * x^{3/2}}{\sqrt{3}}\)

Step 1: Break down the numerator and denominator under the root

Break down the \(150 x^{4}\) and \(3 x\) using the property of square roots, \(\sqrt{ab} = \sqrt{a} * \sqrt{b}\). Thus \(150 x^{4}\) breaks down to \(\sqrt{150} * \sqrt{x^{4}}\) and \(3 x\) breaks down to \(\sqrt{3} * \sqrt{x}\). Rewrite the expression.

Step 2: Simplify the numerator

\(\sqrt{150} * \sqrt{x^{4}}\) can be further simplified as \(5\sqrt{6} * x^{2}\). This is because \(\sqrt{150} = \sqrt{25} * \sqrt{6} = 5\sqrt{6}\), and \(\sqrt{x^{4}} = x^{2}\) because \((x^{2})^2 = x^{4}\). Substitute the simplified expression back into the original equation.

Step 3: Simplify the denominator

The expression \(\sqrt{3} * \sqrt{x}\) simplifies to \(\sqrt{3} * x^{1/2}\), because the square root of a variable is the variable raised to the power of 1/2. Substitute back into the equation.

Step 4: Simplify the expression fully

After simplifying the numerator and denominator, the equation becomes \(\frac{5\sqrt{6} * x^{2}}{\sqrt{3} * x^{1/2}}\). You can simplify this further by dividing both numerator and denominator by \(x^{1/2}\) which results in \(\frac{5\sqrt{6} * x^{3/2}}{\sqrt{3}}\).

Key Concepts

Understanding Square Roots

Square roots are the opposite of squaring a number. If you take a number and multiply it by itself, you find its square. The square root is simply the number that can be multiplied by itself to get the original number. For example, the square of 4 is 16 because 4 times 4 equals 16. Therefore, the square root of 16 is 4.In algebra, square roots are often used to simplify expressions by breaking down numbers and variables into more manageable parts. For example, the expression \(\sqrt{150}\) can be broken down into \(\sqrt{25} \times \sqrt{6}\), because 150 can be factored into 25 and 6. Since the square root of 25 is 5, we simplify \(\sqrt{150}\) to \(5\sqrt{6}\).

• Remember that \(\sqrt{ab} = \sqrt{a} \times \sqrt{b}\), which allows you to handle more complicated square roots easily.
• Variables under a square root follow the same rule: \(\sqrt{x^4} = x^2\) because raising \(x^2\) to the power of 2 gives \(x^4\).

Simplifying Expressions with Fractions

Simplifying expressions involves reducing them to their simplest form, making them easier to understand and work with. When dealing with fractions, simplifying often means breaking down both the numerator and the denominator into their simplest forms.In the square root fraction \(\frac{\sqrt{150x^4}}{\sqrt{3x}}\), we can apply the property \(\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}\) to simplify the expression. By simplifying each part of the expression separately, you can reduce the complexity of the fraction.

• Simplify the numerator by factoring numbers and variables under the square root.
• Simplify the denominator using the same method.

In this instance, break down each square root: \(150x^4\) becomes \(5\sqrt{6}x^2\) and \(3x\) becomes \(\sqrt{3}x^{1/2}\). By matching and simplifying common terms, the expression becomes easier to manage.

The Art of Algebraic Simplification

Algebraic simplification is the process of reducing an expression to its most efficient form by combining like terms and eliminating unnecessary parts. This process helps in making calculations easier and results more understandable.Using the quotient rule for simplifying fractions is an important tool in algebraic simplification. In the given expression, \(\frac{5\sqrt{6}x^2}{\sqrt{3}x^{1/2}}\), notice the variables \(x^2\) in the numerator and \(x^{1/2}\) in the denominator.

• Apply the quotient rule \(\frac{a^m}{a^n} = a^{m-n}\) to simplify fractions with similar bases.
• This gives \(x^{2 - 1/2} = x^{3/2}\).

After simplifying the variables, the expression refines to \(\frac{5\sqrt{6}x^{3/2}}{\sqrt{3}}\). Similarly, when simplifying coefficients, divide 5 by a matching factor in the denominator, completing the simplification process.