Problem 66 - Intermediate Algebra

Question

Multiply. $$ \left(y^{1 / 2}+5\right)\left(y^{1 / 2}+5\right) $$

Step-by-Step Solution

Answer

The product is \(y + 10y^{1/2} + 25\).

Step 1: Understand the Problem

We are asked to multiply the expression \( (y^{1/2} + 5) \) by itself. This is equivalent to evaluating \( (y^{1/2}+5)^2 \).

Step 2: Apply the Formula for Squaring a Binomial

Recall the formula \((a+b)^2 = a^2 + 2ab + b^2\). We can apply this formula here with \( a = y^{1/2} \) and \( b = 5 \).

Step 3: Calculate \( a^2 \)

Calculate \((y^{1/2})^2\). Since squaring a square root cancels the square root, we have \((y^{1/2})^2 = y^{1} = y\).

Step 4: Calculate \( 2ab \)

Calculate \(2 \cdot y^{1/2} \cdot 5\). This is equal to \(10y^{1/2}\) since multiplication is associative and commutative.

Step 5: Calculate \( b^2 \)

Calculate \(5^2\), which is \(25\).

Step 6: Combine the Results

Combine all the results from the prior steps to obtain the final expanded form: \((y^{1/2} + 5)^2 = y + 10y^{1/2} + 25\).

Key Concepts

Squaring a Binomial

Squaring a binomial is a vital technique in algebra, where you take a binomial expression, such as \((a + b)\), and multiply it by itself. The result is a quadratic expression. When we have \((a + b)^2\), it's crucial to remember the formula:

• \(a^2 + 2ab + b^2\)

This formula effectively expands the squared binomial by calculating the square of each term and twice the product of the two terms. Applying this to our example, \((y^{1/2} + 5)^2\), first, calculate \((y^{1/2})^2\). Since squaring a square root cancels out the root, we simplify this to \(y\). Then, multiply the cross terms: \(2 \cdot y^{1/2} \cdot 5\), which equals \(10y^{1/2}\). Finally, compute \(5^2\) to get \(25\). Combining these, the expanded form is \(y + 10y^{1/2} + 25\). Understanding this process is key to handling more complex algebraic expressions.

Radical Expressions

Radical expressions involve roots, such as square roots, cube roots, etc. In our problem, we deal with \(y^{1/2}\), which is equivalent to the square root of \(y\). Simplifying radical expressions means handling and manipulating these roots in calculations.

• Understand the relationship between radicals and exponents: \(y^{1/2}\) is the same as \(\sqrt{y}\).
• Apply properties of exponents when multiplying or squaring radicals, which helps simplify and combine like terms.

In the given exercise, when we square \(y^{1/2}\), it cancels the root, leaving us with \(y\). It's crucial to be comfortable with operations involving radicals, as they frequently appear in math, especially in algebra and calculus.

Associative Property

The associative property is a fundamental principle in mathematics that tells us how numbers are grouped in an operation does not affect the result. It holds for both addition and multiplication. For example:

• For addition: \((a + b) + c = a + (b + c)\)
• For multiplication: \((a \cdot b) \cdot c = a \cdot (b \cdot c)\)

In our problem, the associative property comes into play when we calculate \(2 \cdot (y^{1/2} \cdot 5)\). With this property, it doesn't matter how we group the numbers, the result is still \(10y^{1/2}\). Recognizing and applying the associative property helps in reorganizing and simplifying calculations, making them less intimidating and more manageable.

Commutative Property

This property highlights that the order of numbers in an operation does not affect the result, applicable to both addition and multiplication. Understanding this is essential for calculations in algebra:

• For addition: \(a + b = b + a\)
• For multiplication: \(a \cdot b = b \cdot a\)

In our multiplication example, \(2 \cdot y^{1/2} \cdot 5\), we can rearrange this due to the commutative property to \(5 \cdot 2 \cdot y^{1/2}\) without changing the result. This flexibility is particularly useful in algebra, allowing us to simplify and manipulate expressions with ease. Embracing the commutative property helps you approach problems with various pathways to solve them.