Problem 77 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{y+2}+y=4 $$

Step-by-Step Solution

Answer

The solution is \( y = 2 \); \( y = 7 \) is extraneous.

Step 1: Isolate the Square Root

Start by isolating the square root term on one side of the equation. Move the linear term, \( y \), to the other side: \[ \sqrt{y+2} = 4 - y \]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation: \[ (\sqrt{y+2})^2 = (4 - y)^2 \] This simplifies to: \[ y + 2 = (4 - y)^2 \]

Step 3: Expand and Simplify

Expand the right side of the equation: \[ y + 2 = 16 - 8y + y^2 \] Rearrange the equation to get all terms on one side: \[ y^2 - 9y + 14 = 0 \]

Step 4: Solve the Quadratic Equation

Use the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -9 \), and \( c = 14 \). Calculate the discriminant: \[ b^2 - 4ac = (-9)^2 - 4 \cdot 1 \cdot 14 = 81 - 56 = 25 \] Thus, the solutions are: \[ y = \frac{9 \pm 5}{2} \] This gives \( y = 7 \) and \( y = 2 \).

Step 5: Verify Solutions for Extraneity

Check each solution by substituting back into the original equation: - For \( y = 7 \): \( \sqrt{7+2} + 7 = 3 + 7 = 10 \), which does not equal 4. - For \( y = 2 \): \( \sqrt{2+2} + 2 = 2 + 2 = 4 \), which equals 4. Thus, \( y = 7 \) is extraneous, and we cross it out.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving equations of the form \( ax^2 + bx + c = 0 \). It is especially useful when the equation is not easily factorable. The formula is given by:

• \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In this exercise, after isolating and squaring the radical equation, we were left with a quadratic equation: \( y^2 - 9y + 14 = 0 \). Here:

• \( a = 1 \)
• \( b = -9 \)
• \( c = 14 \)

Calculating the discriminant, \( b^2 - 4ac \), helps determine the number of solutions. It's the part under the square root in the quadratic formula. A positive discriminant indicates two real solutions, a zero discriminant indicates one real solution, and a negative one means no real solutions. For our case, the discriminant is 25, which is positive.
This calculation leads to two potential solutions: \( y = 7 \) and \( y = 2 \). These results need further verification to avoid keeping any extraneous solutions, an often necessary step in solving such equations.

Extraneous Solutions

Extraneous solutions are solutions that arise from the process of solving an equation but are not valid solutions to the original equation.
They often appear when squaring both sides of an equation, as this can introduce "false positives."
In our equation \( \sqrt{y+2} + y = 4 \), after solving the quadratic, we tested both potential solutions against the original equation. This is a crucial step:

• For \( y = 7 \): When substituted back, the equation \( \sqrt{7+2} + 7 = 3 + 7 = 10 \) does not satisfy the original equation \( = 4 \). Hence, \( y = 7 \) is extraneous.
• For \( y = 2 \): Substituting gives \( \sqrt{2+2} + 2 = 2 + 2 = 4 \), which indeed satisfies the equation. Thus, \( y = 2 \) is a valid solution.

To avoid confusion, it is always recommended to verify solutions by plugging them back into the original equation. This practice filters out any extraneous solutions, ensuring that only valid answers remain.

Isolate the Radical

Isolating the radical is a crucial first step when working with equations involving square roots. It simplifies the equation, making subsequent operations, such as squaring both sides, cleaner and more straightforward.
In our given problem \( \sqrt{y+2} + y = 4 \), we begin by moving all non-radical terms to the opposite side:

• \( \sqrt{y+2} = 4 - y \)

This gives clarity to the next steps, especially when we square both sides to eliminate the square root. This step is essential because working with a simpler algebraic form increases accuracy and helps prevent errors. Once the radical is isolated, the equation can be easily transformed by squaring it, transitioning it into a standard quadratic form.
Remember, always begin solving radical equations by isolating the radical term to achieve a streamlined and error-minimized process.