Problem 2 - Algebra 2 and Trigonometry

Question

Chloe changed the equation \(a^{-2}=36\) to the equation \(\frac{1}{a^{2}}=\frac{1}{36}\) and then took the square root of each side. Will Chloe's solution be correct? Explain why or why not.

Step-by-Step Solution

Answer

Yes, Chloe's solution is correct; she correctly transformed and solved the equation.

Step 1: Identify the Original Equation

The original equation given is \(a^{-2}=36\). This suggests that when \(a\) is raised to the power of \(-2\), it equals 36.

Step 2: Change of Equation Interpretation

Chloe changed the given equation to \(\frac{1}{a^2} = \frac{1}{36}\). This step is based on the property of exponents, where \(a^{-n} = \frac{1}{a^n}\). Thus, Chloe's interpretation and transformation is valid.

Step 3: Take the Square Root of Both Sides

Next, Chloe took the square root of each side of the equation \(\frac{1}{a^2} = \frac{1}{36}\). Therefore, \(\sqrt{\frac{1}{a^2}} = \sqrt{\frac{1}{36}}\). This simplifies to \(\frac{1}{a} = \frac{1}{6}\).

Step 4: Solve for 'a'

From the equation \(\frac{1}{a} = \frac{1}{6}\), multiply both sides by \(a\) and 6 to get \(6 = a\). Thus, Chloe correctly solves for \(a\).

Step 5: Conclusion

Since Chloe correctly applied the properties of exponents and square roots to solve the equation, her solution is correct, and the correct value of \(a\) is 6.

Key Concepts

Understanding Exponents

Exponents are a special way of writing repeated multiplication. In simpler terms, when you have an exponent, you are essentially multiplying a number by itself a certain number of times. For example, in the expression \(2^3\), the 3 is the exponent, telling us to multiply 2 three times: \(2 \times 2 \times 2 = 8\).
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There are special rules for managing exponents, which include dealing with negative exponents. When you see a negative exponent like \(a^{-n}\), it means you take the reciprocal or the inverse of the base raised to the positive of that exponent. This means \(a^{-2} = \frac{1}{a^2}\). This rule was applied in the step where Chloe transformed the original equation \(a^{-2}=36\) to \(\frac{1}{a^2} = \frac{1}{36}\). Such rules help simplify and solve many types of algebraic expressions.

The Basics of Square Roots

The square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 36 is 6, because \(6 \times 6 = 36\). The symbol for square root is \(\sqrt{}\).
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When Chloe decided to take the square root of both sides of the equation \(\frac{1}{a^2} = \frac{1}{36}\), she cleverly employed a basic mathematical property. Taking square roots can simplify expressions, especially when they are squared. After applying the square root, the equation \(\sqrt{\frac{1}{a^2}} = \sqrt{\frac{1}{36}}\) becomes \(\frac{1}{a} = \frac{1}{6}\).
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The application of square roots is crucial in solving equations, especially those involving quadratic terms, where a variable is squared.

Algebraic Manipulation Simplified

Algebraic manipulation involves using mathematical techniques to simplify or solve equations. This can include operations like adding, subtracting, multiplying, dividing, and using exponents and roots effectively.
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In Chloe's solution, once she reached \(\frac{1}{a} = \frac{1}{6}\), she performed a critical step. Specifically, both sides of the equation were manipulated by multiplying by \(a\) and by 6 to solve for \(a\). This resulted in \(6 = a\). This kind of operation requires balancing both sides of the equation, ensuring each operation is applied equally to maintain equality.
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Algebraic manipulation is fundamental in solving not only simple equations but also complex algebraic expressions and systems. Developing these skills is essential for students tackling algebra problems.