Problem 65 - Beginning and Intermediate Algebra

Question

Solve. \sqrt{x-3}+\sqrt{x+2}=5

Step-by-Step Solution

Answer

The solution to the equation \(\sqrt{x-3}+\sqrt{x+2}=5\) is \(x=3\).

Step 1: Isolate the square root term\( \sqrt{x-3}\)

Subtract \(\sqrt{x+2}\) from both sides of the equation: \[\sqrt{x-3} = 5 - \sqrt{x+2}\]

Step 2: Square both sides of the equation

Square both sides to eliminate the square root on the left side: \[(\sqrt{x-3})^2 = (5 - \sqrt{x+2})^2\]

Step 3: Simplify and expand the equation

Simplify the left side and expand the right side using the binomial formula: \[x-3 = (5-\sqrt{x+2})(5-\sqrt{x+2})\] \[x-3 = 25 - 10\sqrt{x+2} + (x+2)\]

Step 4: Combine like terms

Subtract \(x\) and add 3 to both sides, then combine like terms: \[0 = 25 - 10\sqrt{x+2} + 2\] \[-x+3-x = -10\sqrt{x+2} + 27\] \[-10\sqrt{x+2} = -2x + 24\]

Step 5: Isolate the square root term \( \sqrt{x+2}\)

Add 10 and divide by -5 to isolate the square root term on the left side: \[\sqrt{x+2} = x - 5\]

Step 6: Square both sides of the equation

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

Step 7: Simplify and expand the equation

Simplify the left side and expand the right side using the binomial formula: \[x+2 = x^2 - 10x + 25\]

Step 8: Bring all terms to one side and simplify into a quadratic equation

Subtract \(x\) and subtract 2 from both sides to get the quadratic in standard form: \[0 = x^2 - 11x + 23\]

Step 9: Solve the quadratic equation

Factor the quadratic equation: \[0 = (x-8)(x-3)\] Thus, the possible solutions for \(x\) are 8 and 3.

Step 10: Check for extraneous solutions

Plug the values of \(x\) back into the original equation to see if they satisfy the equation: 1. For \(x=8\): \[\sqrt{8-3}+\sqrt{8+2}=5\] \[\sqrt{5}+\sqrt{10}=5\] This is not true, so \(x=8\) is an extraneous solution. 2. For \(x=3\): \[\sqrt{3-3}+\sqrt{3+2}=5\] \[\sqrt{0}+\sqrt{5}=5\] This is true, so \(x=3\) is the valid solution.

Step 11: Final Answer

The solution to the equation \(\sqrt{x-3}+\sqrt{x+2}=5\) is \(x=3\).

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree two, typically written in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants. Quadratics can have zero, one, or two real solutions. To solve them, you can use methods like:

• Factoring
• Using the quadratic formula: \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)
• Completing the square

In the exercise, we rearranged terms to form the quadratic equation \( x^2 - 11x + 23 = 0 \). Solving this needed us to factor the equation to find potential solutions.

Binomial Expansion

Binomial expansion involves expanding an expression raised to a power, typically involving two terms. In the formula \((a - b)^2 = a^2 - 2ab + b^2\), each part of the binomial is expanded to simplify the equation. Use binomial expansion for:

• Expanding expressions like \((x+y)^n\)
• Simplifying polynomial terms
• Breaking down complex expressions in algebra

In the given exercise, we expanded \((5 - \sqrt{x+2})^2\) to simplify and solve the equation. This step helped remove the square root and led us toward a quadratic equation.

Extraneous Solutions

Extraneous solutions are solutions that arise in the process of solving an equation but do not satisfy the original equation. This often happens when both sides of an equation are squared. To detect them:

• Always check solutions by substituting them back into the original equation.
• Be cautious with squaring steps as they might introduce invalid solutions.

In the exercise, squaring both sides led to potential solutions \(x = 8\) and \(x = 3\). Checking both, only \(x = 3\) was valid for the original equation, with \(x = 8\) being extraneous.

Factoring Polynomials

Factoring polynomials involves writing the polynomial as a product of its factors. It is a useful technique to solve quadratic equations by setting each factor equal to zero.When factoring:

• Look for a greatest common factor (GCF) first.
• Use methods like grouping or trial and error for trinomials.
• Ensure that the factors multiply back to the original polynomial.

In the example, we factored \(x^2 - 11x + 23 = 0\) into \((x-8)(x-3)\), which led us to the potential solutions \(x = 8\) and \(x = 3\). This process simplified solving the quadratic equation formed.