Problem 226 - Elementary Algebra

Question

In the following exercises, simplify. $$ 2 \sqrt{98}-4 \sqrt{72} $$

Step-by-Step Solution

Answer

-10 \( \sqrt{2} \)

Step 1 - Simplify the radicands separately

Start by simplifying both radical expressions separately. For \( \sqrt{98} \): \[ \sqrt{98} = \sqrt{2 \times 49} = \sqrt{2 \times 7^2} = 7 \sqrt{2} \] For \( \sqrt{72} \): \[ \sqrt{72} = \sqrt{2 \times 36} = \sqrt{2 \times 6^2} = 6 \sqrt{2} \]

Step 2 - Substitute simplified forms back

Replace the simplified forms back into the original expression: \[ 2 \sqrt{98} - 4 \sqrt{72} = 2 \left( 7 \sqrt{2} \right) - 4 \left( 6 \sqrt{2} \right) \]

Step 3 - Multiply constants with the simplified radicals

Perform the multiplication of the constants with the simplified radicals: \[ 2 \times 7 \sqrt{2} = 14 \sqrt{2} \] \[ 4 \times 6 \sqrt{2} = 24 \sqrt{2} \] So, the expression becomes: \[ 14 \sqrt{2} - 24 \sqrt{2} \]

Step 4 - Combine like terms

Combine the like terms (terms with \(\sqrt{2}\)): \[ 14 \sqrt{2} - 24 \sqrt{2} = (14 - 24) \sqrt{2} = -10 \sqrt{2} \]

Key Concepts

Radical Simplification

Simplifying a radical expression is the process of rewriting it in its simplest form. This often involves breaking down a number under the radical sign (called the radicand) into its prime factors, making it easier to take out any perfect squares.

For example, for the radicand in \(\sqrt{98}\):

• First, find the prime factors of 98: 98 = 2 \times 7^2.
• Then, rewrite the radical: \sqrt{98} = \sqrt{2 \times 7^2}.

Notice that 7 is a perfect square, which we can pull out of the radical: \(\sqrt{2 \times 7^2} = 7 \sqrt{2}\). Similarly, simplify \(\sqrt{72}\) by finding its prime factors and pulling out the perfect squares.

• Prime factors of 72: 72 = 2 \times 6^2.
• Rewriting the radical: \sqrt{72} = \sqrt{2 \times 6^2}

We then get: \(\sqrt{2 \times 6^2} = 6 \sqrt{2}\). Now, both expressions are simplified and ready for further operations.

Combining Like Terms

Combining like terms involves adding or subtracting terms that have the same variables and exponents. In the context of radical expressions, this means combining terms with the same radicand.

After simplifying, our expression from the example is 2(\sqrt{98}) - 4(\icefrac{I=72}I) = 2(7\icefrac{I2}) - 4(6\icefrac{I2}). Notice that both terms have \(\sqrt{2}\). So, we can combine them:

Replace the simplified radicals: \(2(7 \sqrt{2}) - 4(6 \sqrt{2})\).

This simplifies further:

• Multiply constants with radicals: 2 \times 7 \sqrt{2} = 14 \sqrt{2}.
• Similarly, 4 \times 6 \sqrt{2} = 24 \sqrt{2}.

Therefore, the expression becomes 14 \sqrt{2} - 24 \sqrt{2}. Since both terms contain \(\sqrt{2}\), we can combine them:

14 \sqrt{2} - 24 \sqrt{2} = (14 - 24) \sqrt{2} = -10 \sqrt{2}.

Multiplication of Radicals

Multiplying radical expressions involves multiplying the numbers outside the radicals together and the radicands together. This can help straightforwardly combine and simplify expressions.

Let’s consider two radical expressions \(a \sqrt{x}\) and \(b \sqrt{y}\). When multiplied:

• Multiply the constants outside the radicals: \(a \times b\).
• Multiply the radicands inside the radicals: \(\sqrt{x \times y}\).

For instance:

• Given radicals: \(2 \sqrt{3}\) and \(4 \sqrt{5}\).
• Multiply constants: \(2 \times 4 = 8\).
• Multiply radicands: \(\sqrt{3 \times 5} = \sqrt{15}\).

Therefore, \(2 \sqrt{3} \times 4 \sqrt{5} = 8 \sqrt{15}\). Here, both multiplication steps maintain the simplicity of the expression without losing any terms.

Square Roots

Understanding square roots is fundamental to working with radical expressions. A square root of a number is a value that, when multiplied by itself, gives the original number.

For example, the square root of 49 is 7 because 7 \times 7 = 49. Similarly, the square root of a variable squared, say \(x^2\), is \(x\).

When simplifying square roots:

• Identify and pull out any perfect squares from the radicand.
• For instance, \(\sqrt{50} = \sqrt{2 \times 25} = \sqrt{2 \times 5^2} = 5 \sqrt{2}\).

This makes complex expressions easier to handle. In the example, we found perfect square factors in both radicands – simplifying them helped us combine and further manipulate the terms effectively.