Problem 33 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the radical equation. Check your solutions. \(\sqrt{x+1}-3 x=1\)

Step-by-Step Solution

Answer

The real solution of the radical equation \(\sqrt{x+1}-3x=1\) is \(x=0\). This was confirmed by substitution back into the original problem.

Step 1: Isolate the radical expression on one side

First, isolate the radical \(\sqrt{x+1}\) on the left side of the equation. Doing that, you get \(\sqrt{x+1}=3x+1\).

Step 2: Remove the root

Next, square both sides of the equation to eliminate the square root. Squaring \(\sqrt{x+1}=3x+1\) gives \(x+1=(3x+1)^2\).

Step 3: Simplify the Right-Hand Side

Expand the right-hand side to get: \(x+1=9x^2+6x+1\).

Step 4: Reduce to a Quadratic Equation

Subtract \(1+x\) from both sides, resulting in the equation: \(0=9x^2+5x\).

Step 5: Solve the Quadratic Equation

This equation can be simplified by factoring out an 'x'. The equation is now \(x(9x+5)=0\). Now, set each factor equal to zero and solve for 'x'. From that, we get \(x=0\) or \(x=-\frac{5}{9}\).

Step 6: Check solutions in the original equation

By substituting \(x=0\) in \( \sqrt{x+1}-3x=1 \), we get \( \sqrt{0+1}-3*0=1 \), which gives \(1=1\). A true statement, thus \(x=0\) is a solution. For \(x=-\frac{5}{9}\) in \( \sqrt{x+1}-3x=1 \), we get \( \sqrt{(-\frac{5}{9})+1}+3*(-\frac{5}{9})=1 \), which gives \(1 \neq 1\). A false statement, thus \(x=-\frac{5}{9}\) is not a solution.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(aeq0\). In our exercise, we encounter a quadratic equation while solving a radical equation, after removing the square root and simplifying the expression to \(x+1=9x^2+6x+1\). The next step involves rearranging and simplifying this equation to match the standard quadratic form, resulting in \(0 = 9x^2 + 5x\).

Quadratics are prominent in many math problems due to their simple yet universal nature. Solving quadratic equations may involve methods such as factoring, using the quadratic formula, completing the square, or graphing.

In this problem, factoring becomes the most efficient approach. Factoring simplifies the equation into more manageable terms for finding solutions.

Factoring

Factoring means breaking down a complicated expression into smaller, more manageable factors that can be multiplied to give the original equation. In the quadratic equation \(0=9x^2+5x\), we can factor out \(x\) from all terms, resulting in \(x(9x+5)=0\).

Factoring transforms the task of solving a quadratic equation into solving two simpler linear equations:

• Set each factor equal to zero: \(x=0\) or \(9x+5=0\).
• Solve for \(x\) in each of these equations:
  • From \(x=0\), we get one possible solution as \(x=0\).
  • From \(9x+5=0\), solving gives \(x=-\frac{5}{9}\).

By factoring, the equation becomes more straightforward, and potential solutions become clear.

Isolating Terms

Isolating terms is a crucial step when working with equations, particularly when they involve radicals or complex expressions. In the original problem, the equation initially is presented as \(\sqrt{x+1}-3x=1\).

To solve this, we first need to isolate the radical term \(\sqrt{x+1}\). By adding \(3x\) to both sides of the equation, we successfully isolates it to form: \(\sqrt{x+1}=3x+1\).

Isolating the radical allows us to square both sides without complicating the equation. Squaring both sides then leads to eliminating the square root, thereby simplifying the equation to more familiar terms, setting the stage for further manipulation and an easier path to the solution.

Verifying Solutions

Verifying the solutions to equations, especially when dealing with radicals, is vital. Despite the mathematical steps seeming faultless, extraneous solutions may emerge, especially after squaring both sides of equations. This is because squaring can introduce solutions that weren't present initially.

For instance, after solving the quadratic \(x(9x+5)=0\), we obtain possible solutions \(x=0\) and \(x=-\frac{5}{9}\). But, we must verify each in the original equation \(\sqrt{x+1}-3x=1\):

• Substitute \(x=0\) into the original equation: \(\sqrt{0+1}-3\times0=1\), which simplifies to \(1=1\), confirming it's true.
• Substitute \(x=-\frac{5}{9}\) into the original equation: \(\sqrt{-\frac{5}{9}+1}+3\times(-\frac{5}{9})=1\), which does not equate to \(1=1\), thereby proving it's false.

Thus, only \(x=0\) is a valid solution. Verification ensures that our results present real and valid solutions to the problem in its original context.