Problem 78 - Elementary and Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. $$ \sqrt{22 y+86}-y=9 $$

Step-by-Step Solution

Answer

The solutions are \( y = 5 \) and \( y = -1 \); both are valid.

Step 1: Isolate the Square Root

Start by adding \( y \) to both sides of the equation to isolate the square root, resulting in \( \sqrt{22y + 86} = y + 9 \).

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation. This gives \( (\sqrt{22y + 86})^2 = (y + 9)^2 \), leading to \( 22y + 86 = y^2 + 18y + 81 \).

Step 3: Rearrange and Simplify the Equation

Move all terms to one side of the equation to set it to zero: \( y^2 + 18y + 81 - 22y - 86 = 0 \). Simplify this to get \( y^2 - 4y - 5 = 0 \).

Step 4: Factor the Quadratic Equation

Factor the quadratic equation \( y^2 - 4y - 5 = 0 \) to find \( (y - 5)(y + 1) = 0 \).

Step 5: Solve for y

Set each factor equal to zero and solve for \( y \): - \( y - 5 = 0 \) gives \( y = 5 \). - \( y + 1 = 0 \) gives \( y = -1 \).

Step 6: Check for Extraneous Solutions

Check both solutions in the original equation to verify: - For \( y = 5 \): \( \sqrt{22(5) + 86} - 5 = 9 \) simplifies to \( \sqrt{196} - 5 = 9 \), which is \( 14 - 5 = 9 \). This solution works. - For \( y = -1 \): \( \sqrt{22(-1) + 86} - (-1) = 9 \) simplifies to \( \sqrt{64} + 1 = 9 \), which is \( 8 + 1 = 9 \). This solution also works.

Step 7: Conclude the Solutions

Both proposed solutions \( y = 5 \) and \( y = -1 \) are valid as they satisfy the original equation.

Key Concepts

Extraneous Solutions

When solving radical equations, we sometimes encounter solutions that do not satisfy the original equation. These are known as extraneous solutions. They often arise when both sides of an equation are squared, which is a common step when dealing with square roots. Squaring usually expands possible solutions beyond those that fit the original equation. To spot extraneous solutions:

• Solve the equation as you normally would.
• Substitute each resulting solution back into the original equation.
• Ensure that each solution works in the context of the original problem.

Be cautious: just because a value solves the squared equation doesn't mean it solves the original one. Whenever you solve a radical equation, always check for extraneous solutions to avoid arriving at incorrect conclusions.

Factoring Quadratic Equations

Factoring is a crucial technique in algebra, especially when solving quadratic equations. Once you transform a problem into a standard quadratic form, like \( ax^2 + bx + c = 0 \), you can try to express it as a product of two binomials. That's what we call factoring.Consider a typical quadratic, such as the one from the step-by-step solution: \( y^2 - 4y - 5 = 0 \). To factor:

• Find two numbers whose product equals the constant term, \( c \).
• Ensure those same numbers add up to the middle coefficient, \( b \).

In this case, \(-5\) and \(1\) multiply to \(-5\) and add to \(-4\), allowing us to write:

• \( (y - 5)(y + 1) = 0 \)

Solving each factor separately finds the possible solutions for \( y \). Factoring simplifies to a methodical search for pairs that meet these criteria.

Checking Solutions

Once you have proposed solutions, the next step is ensuring they're correct. This involves substituting back into the original equation to verify.For the equation \( \sqrt{22y+86}-y=9 \):

• Check \( y = 5 \): Substitute back, \( \sqrt{22(5) + 86} - 5 = 9 \) simplifies correctly to confirm \( 14 - 5 = 9 \).
• Check \( y = -1 \): Plug in \( \sqrt{22(-1) + 86} - (-1) = 9 \), simplifying to \( 8 + 1 = 9 \), which is correct.

Carefully checking solutions is essential as it ensures that no extraneous solutions have snuck in. This deliberate checking process guarantees the integrity and accuracy of your results and conclusions.