Problem 62 - Intermediate Algebra

Question

Solve each equation. Write all proposed solutions. Cross out those that are extraneous. See Example 8. $$ 1=\sqrt{4 s+5}-\sqrt{2 s+2} $$

Step-by-Step Solution

Answer

The solutions are \( s = 1 \) and \( s = -1 \), both are valid.

Step 1: Isolate a Square Root

Add \( \sqrt{2s+2} \) to both sides of the equation to isolate one of the square roots: \[ \sqrt{4s+5} = 1 + \sqrt{2s+2} \]

Step 2: Square Both Sides

Square both sides to eliminate the square root: \[ (\sqrt{4s+5})^2 = (1 + \sqrt{2s+2})^2 \] This simplifies to:\[ 4s+5 = 1 + 2\sqrt{2s+2} + 2s + 2 \] Combine like terms:\[ 4s + 5 = 2s + 3 + 2\sqrt{2s + 2} \]

Step 3: Isolate the Remaining Square Root

Subtract \(2s+3\) from both sides to isolate the square root: \[ 2s + 2 = 2\sqrt{2s+2} \]

Step 4: Square Both Sides Again

Square both sides of the equation again to eliminate the remaining square root: \[ (2s+2)^2 = (2\sqrt{2s+2})^2 \] This simplifies to:\[ 4s^2 + 8s + 4 = 4(2s+2) \] Expand the right-hand side:\[ 4s^2 + 8s + 4 = 8s + 8 \]

Step 5: Solve for s

Subtract \(8s + 8\) from both sides:\[ 4s^2 + 8s + 4 - 8s - 8 = 0 \] Simplify:\[ 4s^2 - 4 = 0 \] Divide through by 4:\[ s^2 - 1 = 0 \] This factors to:\[ (s-1)(s+1) = 0 \] Thus, \( s = 1 \) or \( s = -1 \).

Step 6: Check for Extraneous Solutions

Substitute \( s = 1 \) into the original equation:\[ 1 = \sqrt{4(1) + 5} - \sqrt{2(1) + 2} \] \[ 1 = \sqrt{9} - \sqrt{4} = 3 - 2 = 1 \] Since both sides equal each other, \( s = 1 \) is a valid solution.Now substitute \( s = -1 \):\[ 1 = \sqrt{4(-1) + 5} - \sqrt{2(-1) + 2} \] \[ 1 = \sqrt{1} - \sqrt{0} = 1 - 0 = 1 \] Since this also holds true, \( s = -1 \) is another valid solution.

Key Concepts

Extraneous Solutions

When solving radical equations, extraneous solutions often appear due to the nature of squaring both sides of an equation. These are solutions that emerge during the manipulation of the equation but do not satisfy the original equation. Recognizing and handling extraneous solutions is crucial in getting the correct answers. Here’s why they occur:

• Squaring both sides can introduce new solutions that weren't there initially.
• When a value makes a denominator zero or introduces a negative under a square root, it’s usually extraneous.

After solving the equation, it’s essential to substitute each solution back into the original equation. This step ensures that both proposed solutions satisfy the initial equation and helps identify if any are extraneous. This verification is vital to avoid accepting incorrect answers as valid.

Isolate Square Root

To effectively solve radical equations, the first step is to isolate the square root on one side of the equation. This step helps simplify subsequent operations. Consider the problem at hand:

• Start by manipulating the equation so that one square root stands alone on one side, while other terms are on the opposite side.
• This usually involves simple arithmetic operations like addition or subtraction.

For example, in the equation given, adding one term to both sides isolates one of the square roots, making it easier to address it in the next steps. Isolating the square root prepares the equation for the squaring step, facilitating an orderly solution process.

Square Both Sides

Once you've isolated a square root, the next significant step is to square both sides of the equation. Squaring removes the square root, transforming the equation into a quadratic or another solvable form.

• This action enables you to eliminate the radical, making the equation easier to solve.
• Be cautious, as this step can introduce extraneous solutions, as previously mentioned.

In our process, after isolating and then squaring, you’ll often find the equation becomes quadratic. Remember to expand and simplify thoroughly to make solving the equation straightforward. Squaring both sides is a powerful tool in dismantling radicals and revealing algebraic structures beneath them.

Valid Solutions

After getting to potential solutions, determining their validity is the finishing touch in solving radical equations. Not all solutions derived from manipulation and simplification steps will be answers for the original equation.

• Substitute each solution back into the original equation.
• Check if both sides of the initial equation hold true with each proposed solution.

This step discards any extraneous solutions, confirming those that genuinely satisfy the equation. As exhibited in the solution process, both potential answers were verified to be valid, giving confidence in the correctness of results. Checking for valid solutions ensures the solutions’ integrity and correctness.