Problem 41 - Algebra 2

Question

Solve. Check for extraneous solution. $$x^{\frac{1}{2}}-(x-5)^{\frac{1}{2}}=2$$

Step-by-Step Solution

Answer

The solution for the equation is \(x = 5 - \sqrt{21}\).

Step 1: Square both sides of the equation

Eliminate the square roots by squaring both sides of the equation.\[(x^{\frac{1}{2}}-(x-5)^{\frac{1}{2}})^2=(2)^2\]This results in:\[x - 2x^{\frac{1}{2}}(x - 5)^{\frac{1}{2}} + (x - 5) = 4\]

Step 2: Isolate Radical Terms

Keep the term with square roots on one side and move rest of the terms to the other to isolate the radical term.\[2x^{\frac{1}{2}}(x - 5)^{\frac{1}{2}} = x - (x - 5) - 4\]Which simplifies to:\[2x^{\frac{1}{2}}(x - 5)^{\frac{1}{2}} = 5\]

Step 3: Square both sides of the equation again

Once again square both sides of the equation to eliminate the square roots.\[(2x^{\frac{1}{2}}(x - 5)^{\frac{1}{2}})^2 = (5)^2\]Simplifying this gives us:\[4x(x - 5) = 25\]Further simplifying this equation results in a quadratic equation:\[4x^2 - 20x - 25 = 0\]

Step 4: Solve the quadratic equation

Use the quadratic formula to solve this equation.\[x = \frac{-(-20) \pm \sqrt{(-20)^2 - 4*(4)*(-25)}}{2*(4)}\]This gives us the roots:\[x = 5 \pm \sqrt{21}\]

Step 5: Check for Extraneous Solutions

Substitute these values into the original equation to check if they are valid or extraneous. Upon checking the solutions in the original equation, we find that \(x = 5 + \sqrt{21}\) is an extraneous solution and \(x = 5 - \sqrt{21}\) is the valid solution.

Key Concepts

Extraneous Solutions

When solving equations, especially radical equations, we sometimes encounter solutions that solve the modified equation but not the original one. These are called extraneous solutions. They often arise when both sides of an equation are squared. This process can sometimes introduce solutions that weren't valid before. Checking your solutions is crucial.

• Always substitute your solutions back into the original equation.
• If the original equation holds true, the solution is valid.
• If not, it’s an extraneous solution and should be discarded.

In our problem, after solving, we ended up with two potential answers: \(x = 5 \pm \sqrt{21}\). Checking these in the original equation showed us that \(x = 5 + \sqrt{21}\) doesn’t satisfy the equation, making it extraneous. Always remember to perform this check to ensure your solutions are accurate.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\). In our solution, after eliminating the square roots, we transformed the problem into a quadratic equation.Solving quadratic equations can be done using different methods such as:

• Factoring
• Using the quadratic formula
• Completing the square

For our problem, we used the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives us the roots of the quadratic equation. It's particularly useful when the equation cannot be factored easily. Make sure to simplify the equation as much as possible before applying the formula for ease of calculation.

Square Roots

Square roots are the opposite operation of squaring a number. In equations, they can introduce complexity, requiring careful handling. In our problem, two square roots were present. When working with square roots:

• You might need to isolate them to simplify your equation.
• Eliminating them often involves squaring both sides of the equation.
• Remember, squaring can introduce extraneous solutions, so always check your final answers.

In this problem, we squared twice to remove all square roots quickly, which helped convert the equation into a more manageable quadratic form.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for unknowns. It’s a fundamental skill in algebra used to isolate variables and simplify complex expressions. In this exercise, the initial step required manipulating the equation to isolate terms with square roots. This required careful subtraction and division, followed by squaring actions. Some useful techniques include:

• Combining like terms
• Moving terms across the equation
• Squaring both sides when dealing with square roots

Through algebraic manipulation, we successfully turned a complicated radical equation into a straightforward quadratic one. Mastering these techniques is invaluable for tackling various algebraic problems efficiently.