Problem 81 - Elementary and Intermediate Algebra

Question

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

Step-by-Step Solution

Answer

The solution to the equation is \(x = 4\).

Step 1: Isolate one of the square roots

Rewrite the equation to have one of the square roots isolated on one side.Start with:\[ 2 = \sqrt{x+5} - \sqrt{x} + 1 \]Subtract 1 from both sides to simplify:\[ 1 = \sqrt{x+5} - \sqrt{x} \]

Step 2: Square both sides

To eliminate the square roots, square both sides of the equation. This step helps us to deal with both square roots comfortably.\[ (1)^2 = (\sqrt{x+5} - \sqrt{x})^2 \]This simplifies to:\[ 1 = (x+5) - 2\sqrt{x+5}\sqrt{x} + x \]So, the equation becomes:\[ 1 = 2x + 5 - 2\sqrt{x+5}\sqrt{x} \]

Step 3: Isolate the remaining square root

Move all terms except the square root to one side of the equation.Subtract \(2x + 5\) from both sides:\[ 1 - 2x - 5 = -2\sqrt{x+5}\sqrt{x} \]This simplifies to:\[ -2x - 4 = -2\sqrt{x+5}\sqrt{x} \]Divide everything by -2:\[ x + 2 = \sqrt{x+5}\sqrt{x} \]

Step 4: Square both sides again

Square both sides again to eliminate the square root:\[ (x + 2)^2 = (\sqrt{x+5}\sqrt{x})^2 \]Now simplify:\[ x^2 + 4x + 4 = (x+5)x \]Which gives:\[ x^2 + 4x + 4 = x^2 + 5x \]

Step 5: Simplify and solve for \(x\)

Now, simplify the equation to find \(x\):Subtract \(x^2\) from both sides:\[ 4x + 4 = 5x \]Subtract \(4x\) from both sides:\[ 4 = x \]

Step 6: Check for extraneous solutions

Substitute \(x = 4\) back into the original equation to verify it works:\[ 2 = \sqrt{4+5} - \sqrt{4} + 1 \]Calculate each term:\[ 2 = \sqrt{9} - 2 + 1 \]Simplify:\[ 2 = 3 - 2 + 1 \]\[ 2 = 2 \]Since both sides of the equation are equal, \(x = 4\) is a valid solution, not extraneous.

Key Concepts

Understanding Square Roots

Square roots are fascinating and unique mathematical operations. A square root is a value that, when multiplied by itself, gives the original number. For example, the square root of 4 is 2, because when 2 is multiplied by itself, it equals 4. The notation used for square roots is a radical symbol,

• For example, \( \sqrt{9} = 3 \) because 3 multiplied by 3 equals 9.
• Square roots find important usage in various fields, including algebraic equations and geometry.

When dealing with equations containing square roots, it often becomes necessary to eliminate the roots to simplify or solve the equation. This is achieved by squaring both sides, which removes the square root symbol. Take care to follow each step, as adding square roots to typical algebraic operations adds complexity that must be managed carefully. This process requires careful balancing and understanding.

Quadratic Equations in Algebra

Quadratic equations are polynomial equations of the second degree. They have the form \( ax^2 + bx + c = 0 \), where \(a\), \(b\), and \(c\) are constants. These equations often involve finding roots (solutions) for the variable, commonly referred to as \(x\). Here’s what you need to know:

• Solutions can be real or complex numbers, determined using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \).
• Quadratic equations are fundamental in algebra and are often encountered in solving real-world problems.
• To solve them, various methods can be employed, such as factoring, completing the square, or utilizing the quadratic formula.

Solving quadratic equations might also involve manipulating the equation’s terms by isolating variables or simplifying expressions. When nested within or alongside square roots, extra caution is advised to ensure accuracy and to avoid transforming the equation incorrectly.

Navigating Extraneous Solutions

Extraneous solutions are potential solutions that arise during the process of solving an equation, but don't satisfy the original equation once substituted back. They often occur when both sides of an equation are squared, inadvertently introducing additional "solutions."

• For instance, when squaring both sides of an equation to eliminate square roots, you might introduce errors or additional solutions not applicable to the original equation.
• To curtail these errors, always substitute proposed solutions back into the original equation to verify their validity.
• Only the solutions that satisfy the original equation without contradiction are valid.

Understanding how and why extraneous solutions occur can prevent confusion. It's crucial to always double-check your work by substituting the found \(x\)-values back into the initial problem. This ensures clarity and correctness, leaving you with only the correct, consistent solutions.