Problem 20 - Algebra A Combined Function

Question

Solve. $$ \sqrt{x+3}+\sqrt{x-5}=3 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{181}{36} \).

Step 1: Isolate one square root

Start by isolating one of the square root terms. Let's take \( \sqrt{x+3} \) on one side of the equation: \[ \sqrt{x+3} = 3 - \sqrt{x-5} \]

Step 2: Square both sides

Square both sides of the equation to eliminate the square root on the left side: \[ (\sqrt{x+3})^2 = (3 - \sqrt{x-5})^2 \] This simplifies to: \[ x + 3 = 9 - 6\sqrt{x-5} + (x - 5) \] Now simplify the expression: \[ x + 3 = x + 4 - 6\sqrt{x-5} \]

Step 3: Simplify and isolate the remaining square root

Subtract \( x \) from both sides to further isolate the square root: \[ 3 - 4 = -6\sqrt{x-5} \] Which simplifies to: \[ -1 = -6\sqrt{x-5} \]

Step 4: Solve for \( \sqrt{x-5} \)

Divide both sides by \(-6\) to solve for the square root: \[ \sqrt{x-5} = \frac{1}{6} \]

Step 5: Square both sides again

Square both sides again to solve for \( x \): \[ x - 5 = \left(\frac{1}{6}\right)^2 \] This results in: \[ x - 5 = \frac{1}{36} \]

Step 6: Solve for \( x \)

Add 5 to both sides to solve for \( x \): \[ x = 5 + \frac{1}{36} \] This simplifies to: \[ x = \frac{181}{36} \]

Step 7: Verify the solution

Substitute \( x = \frac{181}{36} \) back into the original equation to check if it satisfies. Calculate each square root: \[ \sqrt{\frac{181}{36} + 3} \] \[ = \sqrt{\frac{289}{36}} = \frac{17}{6} \] \[ \sqrt{\frac{181}{36} - 5} \] \[ = \sqrt{\frac{1}{36}} = \frac{1}{6} \] Adding these gives: \[ \frac{17}{6} + \frac{1}{6} = \frac{18}{6} = 3 \] Since the original equation is satisfied, \( x = \frac{181}{36} \) is the correct solution.

Key Concepts

Square Roots

Square roots are mathematical operations that allow us to find a number which, when multiplied by itself, gives us the original number. For example, the square root of 9 is 3, because 3 times 3 equals 9. Calculating square roots can sometimes be complex, especially when they are part of an equation.
When dealing with radical equations like \( \sqrt{x+3} + \sqrt{x-5} = 3 \), it is essential to understand how to manipulate and simplify these roots to solve for the variable. This often involves isolating one or more square root terms, squaring both sides of an equation to eliminate the radicals, and simplifying.
Recognize that when you square both sides, you may introduce extraneous solutions. Always verify your solutions in the original equation to ensure they are valid.

Isolating Variables

Isolating variables is a fundamental step in solving equations. It's all about getting the variable you're solving for to one side of the equation by itself. In the context of the provided exercise, we initially isolate one of the square roots to make the equation easier to solve.

• Begin by isolating \( \sqrt{x+3} \) so it stands alone: \( \sqrt{x+3} = 3 - \sqrt{x-5} \).

By isolating a square root, we can then square both sides to eliminate the square root, simplifying the equation and making further algebraic manipulation possible. Speaking more generally, isolating variables by rearranging the equation is often a good first step in uncovering the simplest pathway to the solution.

Equation Verification

Once you solve an equation, especially one involving square roots, it's crucial to verify your solution. This ensures the answer actually holds true in the original equation since squaring can introduce extraneous solutions.
Verification involves substituting the solution back into the original equation and checking if the left side equals the right side. For our problem of \( \sqrt{x+3} + \sqrt{x-5} = 3 \), we need to substitute \( x = \frac{181}{36} \) back into the equation:

• Compute \( \sqrt{\frac{181}{36} + 3} = \frac{17}{6} \)
• Compute \( \sqrt{\frac{181}{36} - 5} = \frac{1}{6} \)
• Verify: \( \frac{17}{6} + \frac{1}{6} = 3 \)

Since both sides match, the solution is correct. Verification reinforces the integrity of your result and ensures complete understanding.

Algebraic Manipulation

Algebraic manipulation involves changing the layout of an equation to simplify or solve it. This includes techniques like distributing, combining like terms, adding, subtracting, multiplying, and dividing both sides of the equation, and more.
For solving the given radical equation, algebraic manipulation occurs several times. Initially, after isolating one square root, you square both sides to get rid of the square root and are left with:

• From \( x + 3 = 9 - 6\sqrt{x-5} + (x - 5) \), simplify to \( x + 3 = x + 4 - 6\sqrt{x-5} \)
• Subtract \( x \) from both sides giving \( 3 - 4 = -6\sqrt{x-5} \)
• Solve further to derive \( \sqrt{x-5} = \frac{1}{6} \)

Each of these steps exemplifies algebraic manipulation and is crucial for simplifying and solving complex problems. Mastering these skills allows for flexible thinking and effective problem-solving in mathematics.