Problem 26 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ x+\sqrt{x+1}=5 $$

Step-by-Step Solution

Answer

The solution set is \( \{ 3 \} \).

Step 1: Isolate the Square Root

First, we need to isolate the square root on one side of the equation. Starting from the original equation: \[ x + \sqrt{x+1} = 5 \] Subtract \( x \) from both sides to isolate the square root: \[ \sqrt{x+1} = 5 - x \]

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation:\[ (\sqrt{x+1})^2 = (5-x)^2 \] This simplifies to: \[ x+1 = (5-x)^2 \] Now, expand the right side:\[ x+1 = 25 - 10x + x^2 \]

Step 3: Rearrange into a Quadratic Equation

Rearrange the equation to bring all terms to one side, setting the equation to zero:\[ x^2 - 10x + 25 - x - 1 = 0 \] Simplify this expression:\[ x^2 - 11x + 24 = 0 \] Thus, we have a quadratic equation: \\( x^2 - 11x + 24 = 0 \)

Step 4: Solve the Quadratic Equation

We will factor the quadratic equation to solve for \( x \). Looking for two numbers that multiply to 24 and add to -11: \( (x - 3)(x - 8) = 0 \)Set each factor to zero and solve for \( x \):- \( x - 3 = 0 \Rightarrow x = 3 \)- \( x - 8 = 0 \Rightarrow x = 8 \)

Step 5: Verify the Solutions

Substitute \( x = 3 \) and \( x = 8 \) back into the original equation to check:- For \( x = 3 \):\[ x + \sqrt{x+1} = 3 + \sqrt{3+1} = 3 + 2 = 5 \] (valid)- For \( x = 8 \):\[ x + \sqrt{x+1} = 8 + \sqrt{8+1} = 8 + 3 = 11 \] (not valid)Only \( x = 3 \) satisfies the original equation.

Step 6: Write the Solution Set

Since only \( x = 3 \) works, the solution set is \( \{ 3 \} \).

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). These equations are called "quadratic" because they contain the term \( x^2 \), which is the square of the variable. Solving quadratic equations is a fundamental skill in algebra that comes in handy in numerous applications across different fields. There are several methods to solve quadratic equations:

• Factoring: Expressing the equation as a product of its factors. We will explore this method in the following sections.
• Quadratic formula: Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions. This formula can solve any quadratic equation.
• Completing the square: Rewriting the equation in perfect square form, facilitating easier solving.

Quadratic equations often have two solutions because they represent parabolas, which can intersect the x-axis at up to two points. A strong understanding of quadratic equations forms a basis for advanced mathematical concepts.

Square Roots

Square roots are an essential concept in mathematics and are commonly written as \( \sqrt{a} \), representing the number that, when multiplied by itself, gives the original number \( a \). For example, \( \sqrt{9} = 3 \) because \( 3 \times 3 = 9 \). When solving equations involving square roots, like \( x + \sqrt{x+1} = 5 \), it's often necessary to manipulate the equation to remove the square root.
To deal with square roots in equations, we can use these strategies:

• Isolation: Move terms around to isolate the square root on one side of the equation, for instance, making \( \sqrt{x+1} = 5 - x \).
• Squaring: Remove the square root by squaring both sides of the equation, ensuring that any square roots are eliminated, simplifying to a potentially solvable form.

It's crucial to check solutions when dealing with square roots since squaring can introduce extraneous solutions.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as a product of simpler expressions or factors. For example, the quadratic equation \( x^2 - 11x + 24 = 0 \) can be factored into \( (x - 3)(x - 8) = 0 \). Factoring is an efficient way to find the roots of an equation because once it's in factored form, the solutions are immediately apparent. This relies on the zero product property which states: If a product of factors is zero, at least one of the factors must be zero.
Here's how factoring works:

• Identify two numbers: These numbers should multiply to give the constant term (here, 24) and add up to the linear coefficient (here, -11).
• Express the equation: Use these numbers to break down the middle term, and then rewrite the equation as the product of two binomials.
• Solve the factors: Set each binomial equal to zero to find the solutions for \( x \). In this case, solving \( x - 3 = 0 \) gives \( x = 3 \), and \( x - 8 = 0 \) gives \( x = 8 \).

Factoring is a powerful technique, especially when the factors are apparent, rendering the solutions quick and straightforward.

Solution Set

A solution set is the set of all values that satisfy a given equation. It's essentially the "answer" to a problem stated in terms of the variable that you're solving for. In solving \( x + \sqrt{x+1} = 5 \), the process led to potential solutions \( x = 3 \) and \( x = 8 \). Yet, not all solutions remain valid after substituting back into the original equation.
Here’s how to ensure that your solution set is correct:

• Substitution: Replace the variable in the original equation with the found solutions to ensure that they satisfy the equation. For \( x = 3 \), the equation holds, but for \( x = 8 \), it does not.
• Consistency Check: Verify that the solutions are consistent with any operations done during solving, such as squaring or isolating terms, which may introduce extraneous solutions.
• Write the set: Once you've verified which solutions are valid, write them as a solution set, like \( \{ 3 \} \), which represents all correct solutions.

Always being thorough with verification ensures that your solution set is accurate and complete.