Problem 20 - Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution is \(x = \frac{181}{36}\), which satisfies the equation.

Step 1: Isolate one square root term

Begin by isolating one of the square root terms in the equation. Let's isolate \(\sqrt{x+3}\). We re-write the equation as \(\sqrt{x + 3} = 3 - \sqrt{x - 5}\).

Step 2: Square both sides to eliminate one square root

Square both sides of the equation to eliminate the square root on the left side. Doing so gives:\[(\sqrt{x+3})^2 = (3 - \sqrt{x-5})^2\]which simplifies to:\[x + 3 = 9 - 6\sqrt{x-5} + (x-5)\]

Step 3: Simplify the equation

Simplify the equation from Step 2:\[x + 3 = 9 - 6\sqrt{x-5} + x - 5\]which simplifies to:\[x + 3 = x + 4 - 6\sqrt{x-5}\]Cancel \(x\) from both sides:\[3 = 4 - 6\sqrt{x-5}\]

Step 4: Isolate the remaining square root

Isolate the remaining square root term by moving other terms to the left side of the equation:\[6\sqrt{x-5} = 4 - 3\]simplifies to:\[6\sqrt{x-5} = 1\]

Step 5: Solve for sqrt term

Solve for \(\sqrt{x-5}\) by dividing both sides by 6:\[\sqrt{x-5} = \frac{1}{6}\]

Step 6: Square both sides again

To solve for \(x\), square both sides of the equation:\[(\sqrt{x-5})^2 = \left(\frac{1}{6}\right)^2\]which simplifies to:\[x - 5 = \frac{1}{36}\]

Step 7: Solve for x

Add 5 to both sides to solve for \(x\):\[x = 5 + \frac{1}{36}\]which can be expressed as:\[x = \frac{180}{36} + \frac{1}{36}\]resulting in:\[x = \frac{181}{36}\]

Step 8: Check the solution

Substitute \(x = \frac{181}{36}\) back into the original equation to verify:Calculate \(\sqrt{\frac{181}{36} + 3}\) and \(\sqrt{\frac{181}{36} - 5}\).Ensure \(\sqrt{x+3} + \sqrt{x-5} = 3\).

Key Concepts

Isolation of Square Roots

One of the key steps in solving radical equations, like \(\sqrt{x+3}+\sqrt{x-5}=3\), involves isolating one of the square root terms. Separating these terms makes it easier to eliminate them in later steps. In our example, we start by isolating \(\sqrt{x+3}\):

• Rewrite the equation as \(\sqrt{x+3}=3-\sqrt{x-5}\).

This step is crucial because it prepares us for the next phase, where we can clear the square root and simplify the equation. Think of it as laying a solid foundation for the solution.

Squaring Both Sides

Squaring both sides of the equation is a powerful technique to remove square roots. Once we have an equation such as \(\sqrt{x+3}=3-\sqrt{x-5}\), we can eliminate the square root on one side by squaring it:

• This gives us \((\sqrt{x+3})^2=(3-\sqrt{x-5})^2\).
• Simplified, it becomes \(x+3=9-6\sqrt{x-5}+(x-5)\).

Remember, squaring eliminates the direct square root, but we have to handle the new elements without losing any terms. Therefore, focus on simplifying afterward, ensuring no term is overlooked.

Equation Solving Steps

After squaring, follow clear solving steps to simplify and solve the equation. Here's a breakdown after isolating square root terms and squaring:

• Simplify the equation to get rid of like terms: \(x+3=x+4-6\sqrt{x-5}\), then cancel \(x\) from both sides to get \(3=4-6\sqrt{x-5}\).
• Isolate the remaining square root: \(6\sqrt{x-5}=1\).
• Divide to solve for \(\sqrt{x-5}\): \(\sqrt{x-5}=\frac{1}{6}\).

Walking through these calculated moves makes solving radical equations more manageable. Take your time to perform each action carefully.

Checking Solutions

It's crucial to check solutions since squaring can introduce extraneous roots. In this case, substitute the found \(x\) back into the original equation to verify its validity:

• The solution \(x=\frac{181}{36}\) must satisfy \(\sqrt{x+3}+\sqrt{x-5}=3\).
• Plug it back and check whether both \(\sqrt{\frac{181}{36}+3}\) and \(\sqrt{\frac{181}{36}-5}\) add up to 3.

Reassessing the solution helps you confirm the accuracy and ensure no steps were overlooked. It's more than just a final check; it's the assurance that our steps and assumptions align with the original equation's requirements.