Problem 25 - Algebra 2 and Trigonometry

Question

In \(3-38,\) solve each equation for the variable, check, and write the solution set. $$ x=1+\sqrt{x+11} $$

Step-by-Step Solution

Answer

The solution set is \( \{ 5 \} \).

Step 1: Isolate the Square Root Term

First, we should isolate the square root term on one side of the equation. The given equation is \( x = 1 + \sqrt{x + 11} \). The square root, \( \sqrt{x+11} \), is already isolated on one side.

Step 2: Eliminate the Square Root

To eliminate the square root, square both sides of the equation. This gives us: \[ (x - 1)^2 = (\sqrt{x + 11})^2 \] which simplifies to: \[ (x - 1)^2 = x + 11 \]

Step 3: Expand the Squared Expression

Expand the left side of the equation: \[ (x - 1)^2 = x^2 - 2x + 1 \] Substituting this back into the equation gives us: \[ x^2 - 2x + 1 = x + 11 \]

Step 4: Rearrange into a Quadratic Equation

Move all terms to one side to form a quadratic equation: \[ x^2 - 2x + 1 - x - 11 = 0 \] Simplifying gives: \[ x^2 - 3x - 10 = 0 \]

Step 5: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic equation is \( x^2 - 3x - 10 = 0 \). We use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -3 \), and \( c = -10 \). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot (-10) = 9 + 40 = 49 \] Plug the values into the formula: \[ x = \frac{-(-3) \pm \sqrt{49}}{2 \cdot 1} = \frac{3 \pm 7}{2} \] This gives two solutions: \( x = 5 \) and \( x = -2 \).

Step 6: Check the Solutions in the Original Equation

Substitute each solution back into the original equation \( x = 1 + \sqrt{x + 11} \):For \( x = 5 \):\[ 5 = 1 + \sqrt{5 + 11} \Rightarrow 5 = 1 + 4 \]. This holds true.For \( x = -2 \):\[ -2 = 1 + \sqrt{-2 + 11} \Rightarrow -2 = 1 + 3 \]. This is not valid.Only \( x = 5 \) is a valid solution.

Step 7: Write the Solution Set

Since only \( x = 5 \) satisfies the original equation, the solution set is \( \{ 5 \} \).

Key Concepts

Square Root

A square root is a special mathematical operation. It looks for a number that, when multiplied by itself, gives the original number you started with. Imagine we have the number 9. The square root of 9 is 3 because 3 times 3 equals 9. We use the symbol \( \sqrt{ } \) to show square roots, just like in the expression \( \sqrt{x+11} \). A neat property of square roots is that they help us figure out how values are structured under certain mathematical conditions.
In equations like \( x = 1 + \sqrt{x + 11} \), understanding the square root helps you find the correct value of \( x \). We rearrange the equation so the square root stands alone. This is important because then we can use algebraic techniques to simplify and solve the equation.

Quadratic Formula

The Quadratic Formula is a powerful tool for solving quadratic equations, which are equations in the form \( ax^2 + bx + c = 0 \). The formula is:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This might look a bit intimidating at first, but once you break it down, it's quite manageable! You simply need to identify the values of \( a \), \( b \), and \( c \) in your quadratic equation, plug them into the formula, and solve for \( x \).
In our example with the equation \( x^2 - 3x - 10 = 0 \), we have \( a = 1 \), \( b = -3 \), and \( c = -10 \). Following the formula steps helps us find potential solutions for \( x \). The formula also uses square roots, connecting back to how these two concepts work closely together.

Solution Set

A Solution Set is simply the list of all solutions that satisfy a given equation. In the context of our problem, after solving the quadratic equation using the quadratic formula, we ended up with two potential solutions: \( x = 5 \) and \( x = -2 \).
However, not all solutions from the quadratic formula will satisfy the original equation. That's why we must verify each solution by substituting it back into the original problem. Once we do this, we notice that \( x = -2 \) does not work within the context of the original equation \( x = 1 + \sqrt{x + 11} \). Hence, our only true solution is \( x = 5 \), and so our solution set is \( \{ 5 \} \).
The solution set essentially communicates which values are truly possible for the given equation considering all mathematical constraints.

Checking Solutions

Checking solutions is a crucial step in solving any equation. Even if you arrive at potential solutions using methods like square roots or the quadratic formula, you should substitute these back into the original equation to verify they really work. Doing this ensures your solutions are consistent with the problem's conditions.
For the equation \( x = 1 + \sqrt{x + 11} \), we check each potential solution:

• Substitute \( x = 5 \): \[ 5 = 1 + \sqrt{5 + 11} \], which simplifies correctly to \( 5 = 5 \).
• Substitute \( x = -2 \): \[ -2 = 1 + \sqrt{-2 + 11} \], which does not hold true as it simplifies to \(-2 = 4 \).

This process confirms \( x = 5 \) is the only correct solution. Checking solutions helps avoid mistakes and verifies the accuracy of your work.