Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. $$ \text { Solve: } \sqrt{x+2}-\sqrt{x-5}=2 $$

Short answer

x = \frac{89}{16}

Step-by-step solution

Isolate one of the square roots

Start by isolating one of the square roots on one side of the equation. Let’s move \ \(\sqrt{x-5}\) to the right side: \[ \sqrt{x+2} = \sqrt{x-5} + 2 \]

Square both sides

Next, square both sides of the equation to remove the square roots: \[ \left(\sqrt{x+2}\right)^2 = \left(\sqrt{x-5} + 2\right)^2 \] \[ x + 2 = (x - 5) + 4\sqrt{x-5} + 4 \]

Simplify the equation

Simplify the equation by combining like terms: \[ x + 2 = x - 1 + 4\sqrt{x-5} \]

Isolate the square root term

Move the x terms to one side of the equation: \[ x + 2 - x + 1 = 4\sqrt{x-5} \] \[ 3 = 4\sqrt{x-5} \]

Solve for the square root

Isolate the square root term by dividing both sides by 4: \[ \sqrt{x-5} = \frac{3}{4} \]

Square both sides again

Square both sides again to solve for x: \[ \left( \sqrt{x-5} \right)^2 = \left( \frac{3}{4} \right)^2 \] \[ x - 5 = \frac{9}{16} \]

Solve for x

Solve for x by adding 5 to both sides: \[ x = \frac{9}{16} + 5 \] \[ x = \frac{9}{16} + \frac{80}{16} \] \[ x = \frac{89}{16} \]

Verify the solution

Substitute \(x = \frac{89}{16}\) into the original equation to verify the solution is correct: \[ \sqrt{\frac{89}{16} + 2} - \sqrt{\frac{89}{16} - 5} = 2 \] This simplifies to \( 2 = 2 \), verifying our solution is correct.

Key concepts

Isolating Square Roots

When solving equations involving square roots, our first goal is to isolate one of the square root terms. This makes it easier to focus on eliminating the square root in subsequent steps. For instance, in the equation given as\( \sqrt{x+2} - \sqrt{x-5} = 2 \), isolating \( \sqrt{x-5} \) involves moving it to the right side:
\[ \sqrt{x+2} = \sqrt{x-5} + 2 \]
Now, we have one square root term isolated, which sets us up for the next important step.
Here are a few tips:

• Always try to isolate the simpler square root if possible.
• Make sure to perform the same operations on both sides to maintain the equality of the equation.

Squaring Both Sides of an Equation

Once a square root has been isolated, the next step is to eliminate it by squaring both sides of the equation. Squaring removes the square root, converting the equation into a more manageable form. Using our example:
\[ \left(\sqrt{x+2}\right)^2 = \left(\sqrt{x-5} + 2\right)^2 \]
Simplifying both sides, we get:
\[ x + 2 = (\sqrt{x-5})^2 + 2\sqrt{x-5} \times 2 + 2^2 \to x + 2 = x - 5 + 4\sqrt{x-5} + 4 \]
Notice that squaring can sometimes introduce extraneous solutions. Always be cautious and verify solutions in the original equation later.
Important points:

• Squaring both sides should correctly handle all terms on both sides of the equation.
• Maintain careful notation to keep track of each term during these manipulations.

Simplifying Equations

After squaring and expanding both sides, the resulting equation can sometimes look complicated. Simplifying the equation is crucial to find a clear path to solving for the variable. In our example, after squaring, simplify to:
\[ x + 2 = x - 1 + 4\sqrt{x-5} \to 3 = 4\sqrt{x-5} \]
Here, subtraction and combining like terms simplify the process.
Next, isolate the square root term, if necessary, and continue to simplify:
\[ \sqrt{x-5} = \frac{3}{4} \]
Once more: square both sides:
\[ (\sqrt{x-5})^2 = \left( \frac{3}{4} \right)^2 \to x - 5 = \frac{9}{16} \to x = \frac{9}{16} + 5 \]
Finally, adjust the equation to match common denominators if needed. This gives:
\[ \frac{89}{16} \]
Confirm the solution by substituting back into the original equation to verify its validity.
To simplify effectively:

• Organize your work step by step to avoid mistakes.
• Double-check each step for arithmetic errors.
• Always verify solutions in the original equation to ensure correctness.