Problem 9 - Algebra 2 and Trigonometry

Question

In \(3-38\) write each expression in simplest form. Variables in the radicand with an even index are non-negative. Variables occurring in the denominator of a fraction are non-zero. $$ \sqrt{250 a^{2}}+\sqrt{10 a^{2}} $$

Step-by-Step Solution

Answer

The expression simplifies to \(6a\sqrt{10}\).

Step 1: Simplify Each Square Root Individually

First, we break down the expressions inside each square root. For \(\sqrt{250 a^2}\), realize that 250 can be factored as \(25 \times 10\), and for any even power of a, say \(a^2\), the square root is \(a\). Therefore, \(\sqrt{250 a^2} = \sqrt{25 \times 10 \times a^2} = 5a\sqrt{10}\). Similarly, for \(\sqrt{10 a^2}\), simplified form is \(a\sqrt{10}\).

Step 2: Combine Like Terms

After simplifying both expressions, we have \(5a\sqrt{10}\) and \(a\sqrt{10}\). Notice that both terms contain \(\sqrt{10}\) as a factor. Treating \(a\) and \(\sqrt{10}\) like variables, these terms are like terms, and thus can be combined: \(5a\sqrt{10} + a\sqrt{10} = (5a + a)\sqrt{10} = 6a\sqrt{10}\).

Key Concepts

Square Roots

Understanding square roots is key to simplifying expressions like \( \sqrt{250 a^2} + \sqrt{10 a^2} \). A square root essentially asks: "What number, when multiplied by itself, gives the original number?" For example, \( \sqrt{25} = 5 \) because \( 5 \times 5 = 25 \). These roots help us simplify expressions involving variables and coefficients.
In algebra, when dealing with square roots of variables like \( a^2 \), the result is \( a \), because \( a \times a = a^2 \). This is crucial when expressions include variables, as simplified forms often look different from their original state. Breaking down the coefficients into factors of perfect squares also helps ease the simplification process.

Factoring

Factoring is a powerful method used to break down numbers or expressions into their base components or 'factors'. In this exercise, we factor numbers within the square roots to simplify the expression.
For example, when simplifying \( \sqrt{250 a^2} \), we can factor 250 into \( 25 \times 10 \). Since \( 25 \) is a perfect square, \( \sqrt{25} = 5 \) can be easily calculated, thus simplifying the inside of the square root. This technique can reveal parts of expressions that can be taken out of the square root, making them easier to handle. Remember that recognizing perfect squares is vital here, as it facilitates further simplification.

Combining Like Terms

Combining like terms is an essential step in simplifying expressions once square roots have been simplified. Terms are considered 'like' when they have the same variables raised to the same power and share the same radical part. The goal is to combine coefficients of these terms.
In our exercise, after simplifying \( \sqrt{250 a^2} \) to \( 5a\sqrt{10} \) and \( \sqrt{10 a^2} \) to \( a\sqrt{10} \), both expressions share \( a\sqrt{10} \). Thus, they are considered like terms. These can be combined by adding their coefficients: \( 5a + a = 6a \). This illustrates how recognizing like terms can streamline solving and further simplify complex algebraic expressions.

Radical Simplification

Radical simplification is the process of making a square root expression as simple as possible. It involves removing perfect squares from under the radical and factoring out common terms. In our case, each square root was simplified first by factoring step by step.
Starting with \( \sqrt{250 a^2} \), we factored out \( 25 \) as a perfect square and moved it out, resulting in \( 5a\sqrt{10} \). For \( \sqrt{10 a^2} \), we directly took \( a^2 \) out as \( a \) and ended with \( a\sqrt{10} \). Such simplification reduces complexity and allows for the effective combination of terms, finally rendering the simpler form \( 6a\sqrt{10} \). These skills are crucial for handling more complex algebraic tasks involving radicals and can significantly reduce computational load.