Problem 36 - Algebra 2 and Trigonometry

Question

In \(3-38\) write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero. $$ \sqrt{4 a b^{2}}-\sqrt{a b^{2}} $$

Step-by-Step Solution

Answer

The simplest form is \( b\sqrt{a} \).

Step 1: Simplify the first square root

Given the expression \( \sqrt{4ab^2} \), we can simplify it. Start by rewriting the radicand as a product of squares: \( \sqrt{4} \times \sqrt{a} \times \sqrt{b^2} \). This simplifies to \( 2b\sqrt{a} \) because \( \sqrt{4} = 2 \), \( \sqrt{a} \) remains unchanged, and \( \sqrt{b^2} = b \). Therefore, \( \sqrt{4ab^2} = 2b\sqrt{a} \).

Step 2: Simplify the second square root

Now consider the expression \( \sqrt{ab^2} \). Here, the radicand can be split into \( \sqrt{a} \times \sqrt{b^2} \). Since \( \sqrt{b^2} = b \), it simplifies to \( b\sqrt{a} \). Therefore, \( \sqrt{ab^2} = b\sqrt{a} \).

Step 3: Subtract the simplified square roots

Now that we have simplified both square roots, the given expression can be rewritten as \( 2b\sqrt{a} - b\sqrt{a} \). This is a straightforward subtraction of like terms. Factor \( b\sqrt{a} \) out of the expression: \( (2b - b)\sqrt{a} \), which simplifies to \( b\sqrt{a} \).

Step 4: Write the final answer

The simplest form of the given expression \( \sqrt{4ab^2} - \sqrt{ab^2} \) is \( b\sqrt{a} \). This is achieved after subtracting and factoring the expression.

Key Concepts

Radicand

The term "radicand" refers to the number or expression inside the square root symbol. It's like the secret ingredient in a recipe, giving the root its base value. In the expression \( \sqrt{4ab^2} \), the radicand is \( 4ab^2 \).
This means it's the part of the expression we need to consider carefully when simplifying the square root.

• To simplify the square root, we look at the radicand and see if it contains perfect squares that can be easily pulled out.
• A perfect square is a number or term that can be squared to return to its original form, like \( 4 \) or \( b^2 \).

In this example, 4 and \( b^2 \) are perfect squares. That's why they could be simplified easily leading to a more "user-friendly" version of the expression. This simplification makes further operations, such as subtraction, much more manageable.

Square Root

A square root is a value that, when multiplied by itself, gives the original number or expression, much like finding one side of a square when you know its area.
It's especially helpful when dealing with expressions involving exponents.
When you encounter \( \sqrt{4ab^2} \), you can break this down by handling each part individually. The goal is to simplify wherever possible:

• \( \sqrt{4} \) becomes 2, as 2 is the number that squares to give 4.
• \( \sqrt{a} \) stays as it is, because a is not a perfect square, or at least not stated as such.
• \( \sqrt{b^2} \) becomes \( b \), since \( b \times b = b^2 \).

This rule of handling each part separately in the radicand allows us to simplify the expression effectively, making square root problems much less intimidating. It’s a methodical approach that works regardless of the complexity of the expression.

Like Terms

"Like terms" are terms whose variables (and their exponents) match exactly, making them combinable through operations such as addition or subtraction.
They are much like apples and apples in a grocery list; you can easily add them together without misunderstanding their identity.
In the expression at hand, \( 2b\sqrt{a} - b\sqrt{a} \), both terms are like terms because they both include the variable part \( b\sqrt{a} \). This allows us to subtract them directly:

• You essentially subtract the coefficients in front of the like terms, similar to what you do in basic arithmetic.
• So, \( 2b\sqrt{a} - b\sqrt{a} \) equals \((2b - b)\sqrt{a} = b\sqrt{a} \).

The ability to efficiently subtract like terms simplifies calculations, minimizing errors and making the overall problem less daunting. Understanding and identifying like terms is a crucial skill in algebraic manipulation, ensuring you can handle complex expressions with finesse.