Spotting Errors In Simplifying Radicals: A Math Challenge
Jamal took on the challenge of simplifying the radical expression $\sqrt{75 x^5 y^6}$ with the conditions $x \geq 0$ and $y \geq 0$. He broke it down like this: $
Now, the question is, can you spot the error in Jamal's work? This isn't just about getting the right answer; it's about understanding the nuances of working with square roots and exponents. Simplifying radicals often involves finding perfect squares within the radicand (the expression under the radical sign) and pulling them out. Jamal correctly identified that 25 is a perfect square ($5^2$), and he correctly pulled the 5 out. He also correctly recognized that $x^4$ is a perfect square ($(x2)2$) and $y^6$ is a perfect square ($(y3)2$). When you take the square root of $x^4$, you get $x^2$, and when you take the square root of $y^6$, you get $y^3$. This is where we need to pay close attention. He wrote $5 x^2 y^2 \sqrt{3 x}$ as his final answer. Let's re-examine each step to see if everything lines up perfectly. The original expression is $
\sqrt{75 x^5 y^6}$. We can rewrite 75 as $25 \cdot 3$. So,
\sqrt{75 x^5 y^6} = \sqrt{25 \cdot 3 \cdot x^5 \cdot y^6}$. Now, let's look at the terms that are perfect squares. 25 is a perfect square. $x^5$ can be written as $x^4 \cdot x$, where $x^4$ is a perfect square ($(x^2)^2$). $y^6$ is also a perfect square ($(y^3)^2$). So, we can rewrite the expression as
\sqrt{25 \cdot x^4 \cdot y^6 \cdot 3 \cdot x}$. Now, let's take the square root of the perfect square terms:
\sqrt{25} = 5$.
\sqrt{x^4} = x^2$.
\sqrt{y^6} = y^3$. Putting these together, we get $5 x^2 y^3$. What's left inside the radical is $3 \cdot x$, which is $3x$. So, the simplified expression should be $5 x^2 y^3 \sqrt{3x}$. Comparing this to Jamal's answer, $5 x^2 y^2 \sqrt{3 x}$, we can see a discrepancy. The exponent on the 'y' term outside the radical is $y^2$ in Jamal's answer, but it should be $y^3$. This is the error. It seems Jamal might have incorrectly taken the square root of $y^6$. He might have divided the exponent by 2, but perhaps made a calculation mistake or a transcription error. Remember, when simplifying a square root, you divide the exponent of any perfect square term by 2 to find the term that comes out of the radical. For $y^6$,
6 \div 2 = 3$, so the term outside the radical should be $y^3$, not $y^2$. It's a common pitfall to make small errors in exponent manipulation, especially when dealing with variables. The conditions $x \geq 0$ and $y \geq 0$ are important because they ensure that we don't have to worry about absolute values when taking the square root of even powers. For example,
\sqrt{x^2} = |x|$. However, since $x \geq 0$, $|x| = x$. Similarly,
\sqrt{y^6} = |y^3|$. Since $y \geq 0$, $y^3 \geq 0$, so $|y^3| = y^3$. This confirms that the simplified term for y should indeed be $y^3$. Jamal's work demonstrates a good understanding of the overall process, but a single slip in applying the exponent rules led to an incorrect final answer. This highlights the importance of careful, step-by-step checking in algebraic simplification. It's easy to overlook a small detail, but in mathematics, those small details can change the entire outcome. Always double-check your exponent arithmetic! Understanding how to simplify radicals is a fundamental skill in algebra, often appearing in geometry problems involving distances and areas, and in higher-level mathematics. The ability to break down a complex expression into simpler parts, identify perfect squares, and correctly apply exponent rules is crucial for success. Jamal's attempt, while flawed, provides a valuable learning opportunity for anyone working with these types of problems. It reinforces the idea that practice and attention to detail are key. The next time you encounter a radical simplification, remember Jamal's case and be extra diligent with those exponents!
