Problem 13 - Contemporary Precalculus

Question

Compute and simplify. $$\left(x^{1 / 2}+y^{1 / 2}\right)\left(x^{1 / 2}-y^{1 / 2}\right)$$

Step-by-Step Solution

Answer

Question: Compute and simplify the product of the given binomial expressions: $(x^{1/2}+y^{1/2})(x^{1/2}-y^{1/2})$ Answer: After computing and simplifying the product, the resulting expression is 0.

Step 1: Apply the Distributive Property

To compute the product, we will start by applying the distributive property. We multiply each term in the first binomial with each term in the second binomial: $$(x^{1 / 2}+y^{1 / 2})(x^{1 / 2}-y^{1 / 2}) = x^{1 / 2} \cdot x^{1 / 2} - x^{1 / 2} \cdot y^{1 / 2} + y^{1 / 2} \cdot x^{1 / 2} - y^{1 / 2} \cdot y^{1 / 2}$$

Step 2: Multiply and Simplify the Terms

We will now multiply and simplify each of the terms obtained from the distributive property: $$x^{1 / 2} \cdot x^{1 / 2} = x^{(1 / 2)+(1 / 2)} = x^1 = x$$ $$x^{1 / 2} \cdot y^{1 / 2} = xy^{1 / 2} / x^{1/2}= xy^{1/2}/x^{1/2}$$ $$y^{1 / 2} \cdot x^{1 / 2} = yx^{1 / 2} / y^{1/2}= yx^{1/2}/y^{1/2}$$ $$y^{1 / 2} \cdot y^{1 / 2} = y^{(1 / 2)+(1 / 2)} = y^1 = y$$

Step 3: Combine and Simplify

Now, we will combine the terms obtained in step 2 and simplify the final expression: $$x - xy^{1/2}/x^{1/2} + yx^{1/2}/y^{1/2} - y$$ The middle terms are equal and opposite in sign, so they cancel each other out: $$x - x + y - y$$ So, the simplified expression is: $$0$$ The product and simplification of the given expression is: $$\left(x^{1 / 2}+y^{1 / 2}\right)\left(x^{1 / 2}-y^{1 / 2}\right) = 0$$

Key Concepts

Exponentiation

Exponentiation is a mathematical operation that raises a number to the power of an exponent. In simpler terms, it tells us how many times to use the number in a multiplication. For example, in the expression \( x^{1/2} \), the exponent is \( \frac{1}{2} \), which represents a square root: \( x^{1/2} = \sqrt{x} \).

• When multiplying terms with the same base, you add their exponents: \( x^{a} \cdot x^{b} = x^{a+b} \).
• For example, \( x^{1/2} \cdot x^{1/2} = x^{1/2+1/2} = x^{1} = x \).
• This is a key concept when multiplying expressions with exponents.

Grasping exponent rules makes it easier to simplify algebraic expressions, as seen in our exercise, where simplifying leads to the cancellation of terms.

Distributive Property

The Distributive Property is a fundamental algebraic concept that allows us to distribute multiplication over addition or subtraction. It can be stated as: \( a(b + c) = ab + ac \).
In the exercise, we use the distributive property to expand the expression \((x^{1 / 2}+y^{1 / 2}) (x^{1 / 2}-y^{1 / 2})\). This involves multiplying each term by every term in the other binomial:

• \( x^{1/2} \cdot x^{1/2} \)
• \( -x^{1/2} \cdot y^{1/2} \)
• \( y^{1/2} \cdot x^{1/2} \)
• \( -y^{1/2} \cdot y^{1/2} \)

This reveals how the distributive property breaks the expression into simpler parts, facilitating easy simplification upon further operations like cancellation.

Algebraic Simplification

Algebraic simplification is the process of making an algebraic expression simpler, often by combining like terms and reducing expressions.
In our context, once the distributive property is applied and each term multiplied, we get expressions like \( x^{1/2} \cdot y^{1/2} - y^{1/2} \cdot x^{1/2} \).

• Here, terms \( xy^{1/2}/x^{1/2} \) and \( yx^{1/2}/y^{1/2} \) are seen to cancel each other out because they are equal but opposite in sign.
• This type of simplification involves identifying terms that can be combined or nullified.
• The result of simplification in our exercise is a zero, showing the powerful impact of algebraic manipulation.

Mastering simplification allows for efficiently handling complex expressions, revealing the essence of algebraic manipulation.