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Chapter 10: Problem 52

Multiply, and then simplify if possible. \(\sqrt{5 y}(\sqrt{y}+\sqrt{5})\)

Short Answer

Expert verified
\( y\sqrt{5} + 5\sqrt{y} \)

Step by step solution

01

Distribute

We need to multiply each term inside the parentheses by the expression outside. Distribute the \(\sqrt{5y}\) to both \(\sqrt{y}\) and \(\sqrt{5}\). This gives us the expression: \[ \sqrt{5y} \cdot \sqrt{y} + \sqrt{5y} \cdot \sqrt{5} \]
02

Simplify the Products

Simplify each term individually. The first term \(\sqrt{5y} \cdot \sqrt{y}\) can be combined as follows: \[ \sqrt{5y} \cdot \sqrt{y} = \sqrt{5y \cdot y} = \sqrt{5y^2} \]The second term \(\sqrt{5y} \cdot \sqrt{5}\) simplifies to: \[ \sqrt{5y} \cdot \sqrt{5} = \sqrt{5 \cdot 5y} = \sqrt{25y} \]
03

Apply Square Root Simplification

Now let's simplify the square roots. For \( \sqrt{5y^2} \), since \( \sqrt{y^2} = y \), we can simplify further: \[ \sqrt{5y^2} = \sqrt{5} \cdot y \]For \( \sqrt{25y} \), since \( \sqrt{25} = 5 \), it simplifies to: \[ \sqrt{25y} = 5 \cdot \sqrt{y} \]
04

Combine the Simplified Terms

Combine the simplified expressions to form the final simplified expression:\[ y\sqrt{5} + 5\sqrt{y} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distribution in Algebra
Distributing in algebra is similar to distributing items in a pool; you spread what's outside the parentheses to each term inside. In algebra, this means taking a term outside the parentheses and multiplying it by each term within them. This exercise begins with the expression \( \sqrt{5y}(\sqrt{y} + \sqrt{5}) \), where we need to distribute \( \sqrt{5y} \) across the terms inside the parentheses.
  • This movement results in two separate multiplications: \( \sqrt{5y} \times \sqrt{y} \) and \( \sqrt{5y} \times \sqrt{5} \).
  • This step essentially sets the stage for deeper operations by preparing terms for further simplification.
Once you've distributed, it's important to make sure your terms are correctly set up for further calculations.
Multiplying Radicals
The process of multiplying radicals follows a straightforward path. When you have two radicals, you can multiply the numbers inside the square roots together. However, this only works smoothly when both radicals have the same index.
  • For example, multiplying \( \sqrt{5y} \) and \( \sqrt{y} \) gives you \( \sqrt{5y \cdot y} = \sqrt{5y^2} \).
  • Similarly, multiplying \( \sqrt{5y} \) by \( \sqrt{5} \) gives \( \sqrt{5 \times 5y} = \sqrt{25y} \).
It's essential to note when multiplying radicals:
  • The radicands (the numbers inside the square roots) need to be combined under a single square root.
  • Be mindful of simplifying opportunities after the multiplication.
Multiplying radicals is like multiplying ordinary numbers, but it offers its simplification challenges, which we'll cover next.
Simplification in Algebra
Algebraic simplification helps make complex expressions easier to understand and work with. After multiplying radicals, the resulting terms are ready for simplification. Simplification can often make equations more manageable.
  • The term \( \sqrt{5y^2} \) simplifies to \( \sqrt{5} \cdot y \) because \( \sqrt{y^2} = y \).
  • The expression \( \sqrt{25y} \) simplifies to \( 5\sqrt{y} \) because \( \sqrt{25} = 5 \).
Simplification in algebra turns complex results into tidier forms, making calculations much simpler and helping to see solutions more clearly.
Square Root Properties
Understanding square root properties is crucial in simplifying expressions effectively. These properties allow you to manipulate and simplify terms involving square roots.
  • The property \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \) allows breaking down the term for easier multiplication or simplification.
  • Additionally, knowing that \( \sqrt{a^2} = a \) (for any non-negative \( a \)) can help eliminate the square root part when you end up with perfect squares.
These properties are powerful tools that enable simplifying expressions like \( y\sqrt{5} + 5\sqrt{y} \), turning an initially daunting expression into something workable and clean. Embracing these properties is key to mastering expressions with square roots.

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Most popular questions from this chapter

Use the properties of exponents to simplify each expression. Write with positive exponents. $$ \frac{\left(m^{2} n\right)^{1 / 4}}{m^{-1 / 2} n^{5 / 8}} $$

Use rational expressions to write as a single radical expression. $$ \sqrt[3]{y} \cdot \sqrt[5]{y^{2}} $$

Use rational exponents to simplify each radical. Assume that all variables represent positive numbers. $$ \sqrt[6]{4} $$

The number of cellular telephone subscriptions in the United States from 1996 through 2006 can be modeled by the function \(f(x)=33.3 x^{45},\) where \(y\) is the number of cellular telephone subscriptions in millions, \(x\) years after \(1996 .\) (Source: Based on data from the Cellular Telecommunications \& Internet Association, \(1994-2000)\) Use this information to answer Exercises. Use this model to estimate the number of cellular telephone subscriptions in the United States in \(2006 .\) Round to the nearest tenth of a million.

Escape velocity is the minimum speed that an object must reach to escape a planet's pull of gravity. Escape velocity \(v\) is given by the equation \(v=\sqrt{\frac{2 G m}{r}},\) where \(m\) is the mass of the planet, \(r\) is its radius, and \(G\) is the universal gravitational constant, which has a value of \(G=6.67 \times 10^{-11} \mathrm{m}^{3} /\) \(\mathrm{kg} \cdot \sec ^{2} .\) The mass of Earth is \(5.97 \times 10^{24} \mathrm{kg}\) and its radius is \(6.37 \times 10^{6} \mathrm{m} .\) Use this information to find the escape velocity for Earth. Round to the nearest whole number. (Source: National Space Science Data Center)
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