Problem 32 - College Algebra and Calculus: An Applied Approach

Question

Find the real solution(s) of the radical equation. Check your solutions. \(x+\sqrt{31-9 x}=5\)

Step-by-Step Solution

Answer

The only solution to the given radical equation is \(x=3\). Extraneous solution \(x=-2\) does not satisfy the original equation.

Step 1: Isolate the Radical

Firstly, one starts with the given equation \(x+\sqrt{31-9x}=5\), which is then rearranged to isolate the radical term. This gives \(\sqrt{31-9x}=5-x\).

Step 2: Square both sides

Then, we square both sides to eliminate the square root, giving \((5-x)^2 = 31-9x\). This simplifies to \(25-10x+x^2 = 31-9x\).

Step 3: Simplify the equation

Next, we simplify the equation, first by shifting all the terms to one side to give \(x^2-x-6=0\). This can then be factored to \((x-3)(x+2)=0\).

Step 4: Solve for \(x\)

Setting each factor equal to zero gives the potential solutions as \(x=3\) and \(x=-2\).

Step 5: Check the solutions

Substituting these values back into the original equation to check: For \(x=3\), we get \(3+\sqrt{31-9(3)}=5\), which simplifies to \(5=5\), a true statement. However, for \(x=-2\), using the same approach, we get \(-2+\sqrt{31-9(-2)}=5\), which simplifies to \(9=5\), a false statement. Therefore, \(x=-2\) is an extraneous solution and must be discarded.

Key Concepts

Isolating Radicals

When dealing with radical equations, the first crucial step is to isolate the radical part of the equation. This means that you want to get the term with the square root by itself on one side of the equation. In our exercise, the equation is given as \(x + \sqrt{31 - 9x} = 5\). To isolate the square root, you need to move \(x\) to the other side of the equation.

• Subtract \(x\) from both sides to find \(\sqrt{31 - 9x} = 5 - x\).
• Ensure that the term inside the square root is complete by itself on one side.

This simplifies your work and sets up the next step well. Isolating the radical ensures that we can focus on solving just the radical part without interference from other terms.

Squaring Both Sides

Once the radical is isolated, it's time to eliminate it by squaring both sides of the equation. This action is critical because it allows you to convert the radical equation into a quadratic form, which is usually easier to solve. Starting from our equation \(\sqrt{31 - 9x} = 5 - x\), we proceed by squaring:

• Square the left side to get \((\sqrt{31 - 9x})^2 = 31 - 9x\).
• Square the right side to get \((5 - x)^2 = 25 - 10x + x^2\).

Now, the equation looks like \(31 - 9x = x^2 - 10x + 25\), a standard quadratic equation. This method is straightforward but emphasizes careful squaring of both sides to avoid mistakes in equation balancing.

Checking Solutions

After solving the quadratic equation, it is important to verify the solutions, as squaring both sides might introduce extraneous solutions. These are solutions that fit the squared equation but not the original one. From our step-by-step, we found potential solutions \(x = 3\) and \(x = -2\). Now, we need to check:

• Substitute \(x = 3\) back into the original equation: \(3 + \sqrt{31 - 9(3)} = 5\), simplifies to \(5 = 5\), verifying a true statement.
• Substitute \(x = -2\) back: \(-2 + \sqrt{31 - 9(-2)} = 5\), simplifies to \(9 = 5\), a false statement.

Thus, \(x = -2\) is an extraneous solution, which needs to be discarded. The final solution is \(x = 3\). Always remember to plug potential solutions back into the original equation to ensure they actually work.