Problem 1 - Algebra 2 and Trigonometry

Question

Ethan said that to solve the equation \((x+3)^{\frac{1}{2}}=5,\) the first step should be to square both sides of the equation. Do you agree with Ethan? Explain why or why not.

Step-by-Step Solution

Answer

Yes, I agree with Ethan; squaring both sides is the correct first step to solve this equation.

Step 1: Understand the Problem

We need to solve the equation \((x+3)^{\frac{1}{2}}=5\). The main goal is to find a value of \(x\) that satisfies this equation. Ethan suggests squaring both sides as the first step to eliminate the square root.

Step 2: Consider Ethan's Method

Ethan proposes starting by squaring both sides of the equation. This means transforming \((x+3)^{\frac{1}{2}}=5\) into \((x+3)^{\frac{1}{2}^2} = 5^2\). Since \((x+3)^{\frac{1}{2}}\) and 5 are equal, squaring both sides is valid and will simplify this expression.

Step 3: Square Both Sides

Square both sides of the equation as suggested: \(\left((x+3)^{\frac{1}{2}}\right)^2 = 5^2\). This simplifies to \(x+3=25\). Squaring is an appropriate way to eliminate the square root present in the original equation.

Step 4: Solve for x

Now, solve the linear equation to find \(x\): \(x + 3 = 25\). Subtract 3 from both sides to isolate \(x\): \(x = 25 - 3\).

Step 5: Find the Solution

Simplify the expression to find \(x\): \(x = 22\). Thus, the solution to the equation is \(x = 22\). Since squaring both sides led directly to the solution, Ethan's approach is correct.

Key Concepts

Squaring Both Sides

Squaring both sides of an equation is a common technique used to remove square roots. When faced with an equation like \((x+3)^{\frac{1}{2}}=5\), squaring both sides means applying the power of two to the entire equation. Thus, the left-hand side becomes \((x+3)^{\frac{1}{2}^2}\) and the right-hand side becomes \(5^2\).

When you square \((x+3)^{\frac{1}{2}}\), you essentially undo the square root, leaving you with the expression \(x+3\). On the other side, squaring 5 results in 25. The equation then transforms to \(x+3=25\). This makes it easier to solve for \(x\).

It's essential to ensure that both sides of the equation can still represent equal values after squaring. This method is valid as long as the original equation involves equivalent expressions on both sides. Squaring is powerful because it maintains the balance of the equation while removing the square root.

Eliminating Square Roots

Eliminating square roots from equations is crucial for simplifying and solving them effectively. In the context of the equation \((x+3)^{\frac{1}{2}}=5\), eliminating the square root is the primary goal.

By squaring both sides of the equation, you effectively remove the square root. The step transforms the equation into \(x+3=25\), which no longer has any radical expressions.

• First, take note of the expression under the square root (called the radicand). In this case, it’s \(x+3\).
• Secondly, square both the radicand and the other side of the equation to maintain equality.

Eliminating the square root is particularly useful because it reduces the problem to a simpler form. You should ensure that squaring will not introduce any extraneous solutions by double-checking the domain of the original function.

Linear Equations

Once the square root is eliminated, the equation \(x+3=25\) becomes a straightforward linear equation. Linear equations are equations of the first degree, meaning they have the general form of \(ax + b = c\) where \(a\), \(b\), and \(c\) are constants.

In our reduced equation, the task is to isolate \(x\) to determine its value. You achieve this by performing simple arithmetic operations.

• Subtract 3 from both sides to keep \(x\) isolated. This results in \(x = 25 - 3\).
• Simplify the arithmetic to find \(x = 22\).

Linear equations, like this one, are typically easier to solve than equations with radicals. The process of isolating \(x\) allows you to find a neat and clear solution, highlighting the simplicity of linear forms.

Once simplified, the linear result directly gives you the solution, allowing for quick checks and verifications.

Solution Verification

Solution verification is a critical step after finding a possible solution. This involves substituting the found value back into the original equation to ensure it's correct.

For our equation \((x+3)^{\frac{1}{2}}=5\), we found \(x = 22\). To verify:

• Substitute \(x = 22\) back into the original expression \((22+3)^{\frac{1}{2}}\).
• This simplifies to \(25^{\frac{1}{2}}\) which equals 5, confirming the left side matches the right side of the original equation.

If the substitution checks out, the solution is confirmed as correct. If not, you may need to re-evaluate your steps to ensure no errors creep into the process.

Solution verification safeguards against mistakes, providing assurance that the solutive paths taken are valid and the results are trustworthy. It's not just about finding numbers; it's about reinforcing the logic and steps used to arrive at those numbers.