Problem 27 - College Algebra Essentials

Question

Check all proposed solutions. $$ \sqrt{2 x+3}+\sqrt{x-2}=2 $$

Step-by-Step Solution

Answer

The solution to the equation \( \sqrt{2 x+3}+\sqrt{x-2} = 2 \) is \(x = 2\).

Step 1: Checking the existence conditions

For the square roots to be defined, we must have \(2x+3 \geq 0\) and \(x-2 \geq 0\), hence, \(x \geq -\frac{3}{2}\) and \(x \geq 2\). Combining these conditions, we find that \(x ≥ 2\). This will be the domain.

Step 2: Isolating one of the radicals

We can isolate one of the radicals by subtracting \( \sqrt{x-2}\) from both sides of the equation. So we get \( \sqrt{2x+3} = 2 - \sqrt{x-2} \).

Step 3: Squaring to eliminate the radical

Square both sides to eliminate the radical: \((\sqrt{2x+3})^2 = (2 - \sqrt{x-2})^2 => 2x+3 = 4 - 4\sqrt{x-2} + (x-2)\). Simplifying right side we get \(2x+3 = 2 + x - 4\sqrt{x-2}\).

Step 4: Isolate the second radical

To isolate the second radical, we re-arrange the equation from step 3 to move non-radical terms to one side \(4\sqrt{x-2} = x+1\).

Step 5: Squaring to eliminate the second radical

Square both sides of the result from step 4 to eliminate the second radical: \( (4\sqrt{x-2})^2 = (x+1)^2 \). This simplifies to \(16(x-2) = x^2 + 2x + 1\), and further simplifying gives the quadratic equation \(x^2 - 14x - 31 = 0\).

Step 6: Solving the Quadratic Equation

Analyze this as quadratic equation in form of \(ax^2 +bx +c = 0\). Find solutions as \(x = \frac{-b±\sqrt{b^2-4ac}}{2a}\). Solving with \(a=1\), \(b=-14\), and \(c=-31\), we get \(x_1 = 2\), \(x_2 = -15\).

Step 7: Check the solutions

Check the solutions \(x = 2\) and \(x = -15\) in the original equation. However, since \(x = -15\) is outside of our original domain (from step 1), we reject this solution. So the only solution is \(x = 2\).

Key Concepts

Domain of a Function

When solving radical equations, finding the domain of a function is a crucial first step. The domain refers to all the possible values of the variable that make the equation defined and solvable. For an expression under a square root to be valid, it must be non-negative, meaning it cannot be less than zero.

In the example exercise, we have two square roots: \( \sqrt{2x+3} \) and \( \sqrt{x-2} \). This means we have two conditions:

• \( 2x + 3 \geq 0 \), which simplifies to \( x \geq -\frac{3}{2} \)
• \( x - 2 \geq 0 \), which simplifies to \( x \geq 2 \)

To find the domain of the entire equation, we take the stricter of these conditions, which is \( x \geq 2 \). Only values of \( x \) that satisfy this inequality should be considered as potential solutions. Understanding the domain ensures we're only working with valid numbers since any solution outside the domain is invalid.

Quadratic Equation

Quadratic equations appear frequently when solving radical equations, especially after squaring both sides to eliminate radicals. A quadratic equation is generally in the form \( ax^2 + bx + c = 0 \). These can be solved using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula provides two potential solutions, which correspond to the \( \pm \) sign in the expression.

In the given exercise, after isolating and then squaring the radicals, the equation \( x^2 - 14x - 31 = 0 \) emerges. Here:

• \( a = 1 \)
• \( b = -14 \)
• \( c = -31 \)

By applying these values into the quadratic formula, we find the roots \( x_1 = 2 \) and \( x_2 = -15 \). It's important to remember both solutions will require verification, especially with respect to the domain.

Verification of Solutions

After finding the potential solutions to a radical equation, it's essential to verify these solutions to ensure they satisfy the original equation. This step is crucial because not all computed solutions may be valid, especially after squaring steps, which can introduce extraneous solutions.

For our exercise, we have found \( x = 2 \) and \( x = -15 \). However, during verification, we must remember which values lie within the domain determined earlier, \( x \geq 2 \).

• \( x = 2 \) is within the domain and thus needs to be checked against the original equation \( \sqrt{2x+3} + \sqrt{x-2} = 2 \). Substituting \( x = 2 \) fulfills both sides of the equation, confirming it as a valid solution.
• \( x = -15 \), however, is immediately outside the domain \( x \geq 2 \) and can be dismissed without further calculation.

This step of verification ensures that only true solutions that satisfy both the equation and the domain are accepted.