Problem 42 - Elementary Algebra

Question

Simplify each expression by performing the indicated operation. $$ \sqrt{10}(\sqrt{10}-\sqrt{5}) $$

Step-by-Step Solution

Answer

Question: Simplify the expression $\sqrt{10}(\sqrt{10}-\sqrt{5})$. Answer: $10 - \sqrt{50}$

Step 1: Identify the expression given

The given expression is: $$ \sqrt{10}(\sqrt{10}-\sqrt{5}) $$

Step 2: Apply distributive property

Multiply each term within the parentheses by the term outside the parentheses: $$ \sqrt{10} \cdot \sqrt{10} - \sqrt{10} \cdot \sqrt{5} $$

Step 3: Perform the multiplication

Now, multiply each term as indicated: $$ 10 - \sqrt{10 \cdot 5} $$

Step 4: Simplify the expression

Finally, simplify the remaining expression: $$ 10 - \sqrt{50} $$ The simplified expression is: $$ 10 - \sqrt{50} $$

Key Concepts

Distributive Property

Understanding the distributive property is crucial when it comes to simplifying algebraic expressions. In essence, this property allows you to multiply a single term by each term within a parenthesis, effectively 'distributing' the multiplication over addition or subtraction.

For example, if you see an expression like \( a(b + c) \), you can apply the distributive property to get \( ab + ac \). This is what we did in the exercise with \( \sqrt{10}(\sqrt{10}-\sqrt{5}) \), where \( \sqrt{10} \) was distributed to both \( \sqrt{10} \) and \( -\sqrt{5} \) inside the parentheses.

It's important to remember that this property works the same way with subtraction as it does with addition, and it underpins many methods for simplifying more complex expressions.

Square Roots

Square roots are a fundamental concept in algebra that involve finding a number which, when multiplied by itself, gives the original number. For instance, the square root of 25 is 5 because \( 5 \times 5 = 25 \).

In our exercise, we used square roots in the form of \( \sqrt{10} \) and found that squaring it—multiplying \( \sqrt{10} \) by itself—gives us 10. Understanding how to work with square roots is essential when dealing with radical expressions, and it's often a matter of identifying and applying square numbers to simplify an expression effectively. Remember, the key is to find the number that 'squares' to give the original value under the root sign.

Algebraic Operations

Algebraic operations include the basic arithmetic processes—addition, subtraction, multiplication, and division—applied to algebraic expressions. In algebra, we often work with unknowns represented by variables, but the operations themselves remain consistent.

In our initial problem, after applying distributive property, we performed multiplication and found that \( \sqrt{10} \times \sqrt{10} \) equals 10. And \( \sqrt{10} \times \sqrt{5} \) equals \( \sqrt{50} \), because when you multiply roots, you can multiply the numbers under them. Lastly, we subtracted one expression from the other, which illustrated how these operations can combine to simplify complex expressions.

Radical Expressions

Radical expressions include numbers under a root symbol, with square roots being the most common. Simplifying radical expressions often involves applying the properties of square roots, such as the fact that \( \sqrt{a} \times \sqrt{a} = a \).

In the solution to our exercise, after applying the distributive property, we ended up with \( 10 - \sqrt{50} \). This is a radical expression because of the presence of \( \sqrt{50} \). Often, you'll want to simplify such expressions by finding perfect squares within the radicand—the number under the root. In this case, 50 can be expressed as \( 25 \times 2 \), which means the expression could be further simplified by recognizing that \( \sqrt{25} \) is a perfect square (5), leading to \( 10 - 5\sqrt{2} \). Knowing how to break down and simplify radical expressions is a valuable skill for tackling more advanced algebra problems.