Solving $a√u = 4$ Step-by-Step Guide

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In this article, we will delve into the process of solving the equation au=4a\sqrt{u} = 4 for the variable uu. This is a fundamental algebraic problem that involves isolating the variable and applying appropriate mathematical operations. We will break down the solution step-by-step, ensuring clarity and understanding. This problem often arises in various mathematical contexts, including algebra, calculus, and physics, making it a crucial skill to master. By the end of this guide, you will have a comprehensive understanding of how to solve this type of equation and be well-equipped to tackle similar problems.

Understanding the Equation

To effectively solve for uu in the equation au=4a\sqrt{u} = 4, it is vital to first understand the components and structure of the equation. The equation involves a variable uu under a square root, a coefficient aa, and a constant 44. The goal is to isolate uu on one side of the equation. This involves undoing the operations that are applied to uu, such as the square root and multiplication by aa. Understanding the domain of the variable is also important; since uu is under a square root, it must be non-negative (u0u \geq 0). Additionally, we must consider the coefficient aa. If aa is zero, the equation becomes 0=40 = 4, which has no solution. Therefore, we assume that aa is a non-zero constant. This understanding forms the basis for the subsequent steps in solving the equation.

Step-by-Step Solution

1. Isolate the Square Root

The first crucial step in solving the equation au=4a\sqrt{u} = 4 for uu involves isolating the square root term. This is achieved by dividing both sides of the equation by the coefficient aa. As we discussed earlier, it is essential to assume that aa is not equal to zero; otherwise, the division would be undefined, and the equation would not have a solution. Dividing both sides by aa gives us:

aua=4a\frac{a\sqrt{u}}{a} = \frac{4}{a}

This simplifies to:

u=4a\sqrt{u} = \frac{4}{a}

Now, the square root term is isolated on one side of the equation, which sets the stage for the next step in solving for uu. This isolation is a fundamental technique in solving algebraic equations, allowing us to focus on the remaining operations needed to find the variable's value.

2. Square Both Sides

Having isolated the square root term, the next step to solve for uu in the equation u=4a\sqrt{u} = \frac{4}{a} is to eliminate the square root. This is achieved by squaring both sides of the equation. Squaring both sides ensures that we maintain the equality while removing the square root. This operation gives us:

(u)2=(4a)2(\sqrt{u})^2 = \left(\frac{4}{a}\right)^2

When we square the square root of uu, we get uu. On the other side, we square the fraction 4a\frac{4}{a}, which means squaring both the numerator and the denominator:

u=42a2u = \frac{4^2}{a^2}

This simplifies to:

u=16a2u = \frac{16}{a^2}

This step is critical because it directly eliminates the square root, bringing us closer to solving for uu. Squaring both sides is a common technique in algebra when dealing with square roots and other radical expressions.

3. The Solution for u

After squaring both sides of the equation, we have arrived at the solution for uu. The equation u=16a2u = \frac{16}{a^2} expresses uu in terms of aa. This means that the value of uu depends on the value of aa. For any non-zero value of aa, we can find a corresponding value for uu. However, it is important to remember the initial condition that uu must be non-negative because it was under a square root in the original equation. Since a2a^2 is always positive for any non-zero aa, and 16 is positive, the fraction 16a2\frac{16}{a^2} will always be positive. Thus, the solution u=16a2u = \frac{16}{a^2} is valid for all non-zero values of aa.

4. Verification of the Solution

To ensure the correctness of our solution, it is crucial to verify it by substituting the value of uu back into the original equation. This step helps to confirm that our algebraic manipulations were accurate and that the solution satisfies the given equation. Substituting u=16a2u = \frac{16}{a^2} into the original equation au=4a\sqrt{u} = 4, we get:

a16a2=4a\sqrt{\frac{16}{a^2}} = 4

We can simplify the square root term:

16a2=16a2=4a\sqrt{\frac{16}{a^2}} = \frac{\sqrt{16}}{\sqrt{a^2}} = \frac{4}{|a|}

So the equation becomes:

a4a=4a \cdot \frac{4}{|a|} = 4

If aa is positive, then a=a|a| = a, and the equation simplifies to:

a4a=4a \cdot \frac{4}{a} = 4

4=44 = 4

This confirms that the solution is correct for positive values of aa. If aa is negative, then a=a|a| = -a, and the equation becomes:

a4a=4a \cdot \frac{4}{-a} = 4

4=4-4 = 4

This is not true, which indicates that our solution is valid only for positive values of aa. Therefore, the verified solution is u=16a2u = \frac{16}{a^2} for a>0a > 0.

Common Mistakes to Avoid

When solving equations involving square roots, several common mistakes can lead to incorrect solutions. Being aware of these pitfalls can help you avoid them and ensure accurate results. One frequent error is forgetting to consider the domain of the variable under the square root. In the equation au=4a\sqrt{u} = 4, uu must be non-negative because the square root of a negative number is not a real number. Another common mistake is incorrectly applying the squaring operation. When squaring both sides of an equation, it is crucial to square the entire side, not just individual terms. For example, if the equation were u+a=4\sqrt{u} + a = 4, squaring both sides would result in (u+a)2=16(\sqrt{u} + a)^2 = 16, which expands to u+2au+a2=16u + 2a\sqrt{u} + a^2 = 16, not u+a2=16u + a^2 = 16. Additionally, failing to check the solution by substituting it back into the original equation can lead to accepting extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. By being mindful of these common mistakes, you can improve your accuracy in solving equations with square roots.

Conclusion

In conclusion, we have successfully solved the equation au=4a\sqrt{u} = 4 for uu, arriving at the solution u=16a2u = \frac{16}{a^2}. This process involved several key steps, including isolating the square root term, squaring both sides of the equation, and verifying the solution. We also highlighted the importance of understanding the domain of the variable and avoiding common mistakes such as incorrect squaring and neglecting to check for extraneous solutions. Mastering the techniques demonstrated in this guide will not only enable you to solve similar equations but also enhance your problem-solving skills in mathematics and related fields. Remember, practice is key to solidifying your understanding and building confidence in your ability to tackle algebraic challenges. By consistently applying these methods, you can become proficient in solving equations involving square roots and other mathematical expressions.