Problem 49 - Intermediate Algebra

Question

Solve. \(\sqrt{y+3}-\sqrt{y-3}=1\)

Step-by-Step Solution

Answer

The solution is \(y = \frac{37}{4}\).

Step 1: Isolate a Square Root

Set the equation so that one of the square root terms is isolated. Starting with the given equation \(\sqrt{y+3} - \sqrt{y-3}=1\), isolate \(\sqrt{y+3}\):\[\sqrt{y+3} = \sqrt{y-3} + 1\]

Step 2: Square Both Sides

Square both sides of the equation to eliminate the square root on the left side:\[(\sqrt{y+3})^2 = (\sqrt{y-3} + 1)^2\]This simplifies to:\[y+3 = (y-3) + 2\sqrt{y-3} + 1\]

Step 3: Simplify the Equation

Simplify the right side by combining like terms:\[y+3 = y - 2 + 2\sqrt{y-3}\]This can be rewritten as:\[3 + 2 = 2\sqrt{y-3}\]Further simplify to get:\[5 = 2\sqrt{y-3}\]

Step 4: Isolate and Solve for the Second Root

Divide both sides of the equation by 2 to solve for \(\sqrt{y-3}\):\[\frac{5}{2} = \sqrt{y-3}\]

Step 5: Square Both Sides Again

Square both sides of the equation to solve for \(y\):\[\left(\frac{5}{2}\right)^2 = (\sqrt{y-3})^2\]This simplifies to:\[\frac{25}{4} = y - 3\]

Step 6: Solve for y

Add 3 to both sides to find the value of \(y\):\[\frac{25}{4} + 3 = y\]Convert 3 to a fraction \(\frac{12}{4}\), and add:\[\frac{25}{4} + \frac{12}{4} = \frac{37}{4}\]Thus, \(y = \frac{37}{4}\).

Step 7: Verify the Solution

Substitute \(y = \frac{37}{4}\) back into the original equation to verify:\[\sqrt{\frac{37}{4} + 3} - \sqrt{\frac{37}{4} - 3} = 1\]Calculate each square root and check if both sides are equal.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in solving equations, especially complex ones like those involving radicals. In this exercise, we begin by manipulating the original equation with square roots to make it easier to solve. The given equation is \(\sqrt{y+3} - \sqrt{y-3} = 1\). To start, we focus on isolating one of the square root expressions. This often involves adding or subtracting terms from both sides of the equation. By moving \(-\sqrt{y-3}\) to the other side, we achieve the intermediate result \(\sqrt{y+3} = \sqrt{y-3} + 1\).This initial step is crucial as it sets up the equation for further actions, such as squaring both sides, which we will discuss later. Remember, the clearer the equation, the easier it becomes to work with!

Verification of Solutions

Verification of solutions is a critical step after solving an equation. It's important to ensure that the solutions derived actually satisfy the original equation, especially after algebraic manipulation. This is because operations like squaring both sides can introduce extraneous solutions that don't fit the original problem.For the equation \(\sqrt{y+3} - \sqrt{y-3} = 1\), we found \(y = \frac{37}{4}\) as a solution. To verify this solution, we substitute \(y = \frac{37}{4}\) back into the original equation, checking if both sides equal.Perform the calculations:

• Calculate \(\sqrt{\frac{37}{4} + 3}\)
• Calculate \(\sqrt{\frac{37}{4} - 3}\)

If both calculations satisfy the equation (both sides balance), then \(\frac{37}{4}\) is a valid solution. Always remember, verification is the final seal of approval for your solution!

Squaring Both Sides

In solving radical equations, a common technique is squaring both sides of the equation to eliminate square roots. This makes the equation easier to handle. In our exercise, after isolating one square root expression, we proceed by squaring both sides:\[(\sqrt{y+3})^2 = (\sqrt{y-3} + 1)^2\]Squaring the left side removes the square root, giving us \(y+3\). On the right side, we employ the binomial square formula:\((a + b)^2 = a^2 + 2ab + b^2\)Here, \(a = \sqrt{y-3}\) and \(b = 1\), leading to:\[(y-3) + 2\sqrt{y-3} + 1\]This step transforms our equation into one without radicals, making it much more straightforward to manage. Be mindful, though, as squaring can sometimes introduce solutions that weren't in the original equation!

Isolating Variables

Isolating variables is a key technique in solving equations, especially important when handling equations with radicals. Once we eliminate square roots by squaring both sides, our focus shifts towards isolating the variable itself.After squaring in our example, we get:\[y + 3 = y - 2 + 2\sqrt{y-3}\]Our task is to isolate \(\sqrt{y-3}\) to further progress. Simplify the equation by moving terms independently:

• Combine like terms to simplify the equation.
• Reorganize it to isolate the term with the variable.

We find:\[5 = 2\sqrt{y-3}\]Now, we solve by dividing both sides by 2, achieving \(\sqrt{y-3} = \frac{5}{2}\). This isolation readies the problem for further squaring and ultimate discovery of \(y\). This skill is indispensable for any subsequent algebraic operations!