Problem 55 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$3-\sqrt{x}=\sqrt{2 \sqrt{x}-3}$$

Step-by-Step Solution

Answer

x = 6

Step 1: Isolate the Square Root Terms

First, move the \(-\sqrt{x}\) term to the other side of the equation to isolate the square root expressions.\[ 3 - \sqrt{x} = \sqrt{2\sqrt{x}-3} \] becomes \[ 3 = \sqrt{x} + \sqrt{2\sqrt{x} - 3}. \]

Step 2: Square Both Sides

Next, square both sides of the equation to eliminate the square roots. \[ \left(3\right)^2 = \left(\sqrt{x} + \sqrt{2\sqrt{x} - 3}\right)^2 \] Simplifies to: \[ 9 = x + 2\sqrt{x} \sqrt{2\sqrt{x} - 3} + 2\sqrt{x} - 3. \]

Step 3: Isolate the Remaining Square Root

Simplify and isolate the remaining square root term: \[ 9 + 3 = x + 2\sqrt{x} \sqrt{2\sqrt{x} - 3} + 2\sqrt{x}, \] which further simplifies to: \[ 12 = x + 2\sqrt{4x - 6}. \]

Step 4: Square Both Sides Again

To eliminate the remaining square root, square both sides once more: \[ \left(12\right)^2 = \left(x + 2\sqrt{4\sqrt{x} - 6}\right)^2 \] Simplifies to: \[ 144 = x^2 + 4x + 4(4x - 6). \]

Step 5: Solve the Quadratic Equation

Simplify and solve for \(x\): \[ 144 = x^2 + 4x + 16x - 24 \] Combine like terms: \[x^2 + 20x - 168 = 0. \]Factorize the quadratic: \[(x + 28)(x - 6) = 0. \]Set each factor to zero: \[x = -28, \quad x = 6. \] Since \(x\) cannot be negative in the context of the original square root, the valid solution is \[x = 6. \]

Key Concepts

isolating square roots

Isolating the square root is the first step in solving this type of radical equation. It involves moving one of the square root terms to one side, so you can deal with it separately.
In the given equation, we start with:

• 3 - \(\sqrt{x}\) = \(\sqrt{2\sqrt{x}-3}\)

To isolate a square root term, we move \(-\sqrt{x}\) to the other side, which results in:

• 3 = \(\sqrt{x}\) + \(\sqrt{2\sqrt{x}-3}\)

This step is crucial, as it simplifies one side to make the problem easier to handle. Separating the square roots helps highlight the next steps: squaring both sides to eliminate the square roots. Remember, isolating the square roots means getting one of the square roots alone on one side of the equation.

squaring both sides

Once you isolate the square root, the next step is to square both sides of the equation. This process helps eliminate the square roots and simplifies the equation further.
In our case, we begin with:

• 3 = \(\sqrt{x}\) + \(\sqrt{2\sqrt{x} - 3}\)

Squaring both sides:

• (3)^2 = (\sqrt{x} + \(\sqrt{2\sqrt{x} - 3}\))^2
• 9 = x + 2\sqrt{x}\cdot\sqrt{2\sqrt{x} - 3} + (2\sqrt{x} - 3)

This step might result in another radical expression, but it simplifies the structure, making it easier for the next step. Each time you square both sides, ensure to expand and combine like terms correctly. Be prepared for a more straightforward or another round of isolation and squaring.

quadratic equations

After eliminating the square root expressions, you typically end up with a quadratic equation. Quadratic equations have the form \(ax^2 + bx + c = 0\).
Following our prior work: we get

• 9 + 3 = x + 4\sqrt{x}(4\sqrt{x} - 3)
• 144 = x^2 + 4x + 16x - 24
• x^2 + 20x - 168 = 0

Notice how we reorganized and combined terms to simplify the approach better.
Often, quadratic equations can be factored to find the solutions. If not, use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

valid solutions

When solving radical equations, it's essential to check for valid solutions. This ensures that the solutions work within the context of the original equation, especially since squaring can introduce extraneous solutions.
Our factorized quadratic equation is:

• (x + 28)(x - 6) = 0

Setting each factor to zero gives us:

• x = -28
• x = 6

However, negative values often do not apply in the original context of square roots. Therefore, check if the solution makes the original equation true. Here, x = 6 is valid. Reevaluate by substituting back into the original equation to confirm. This step avoids any potential errors, ensuring the solution satisfies all the conditions.