Problem 48 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ x+2 \sqrt{x-7}=10 $$

Step-by-Step Solution

Answer

The only real solution is \(x = 8\).

Step 1: Isolate the Square Root

Start by isolating the square root term. To do this, subtract \(x\) from both sides of the equation: \[ 2 \sqrt{x-7} = 10 - x \]

Step 2: Divide to Simplify

Divide both sides of the equation by 2 to further isolate the square root: \[ \sqrt{x-7} = \frac{10 - x}{2} \]

Step 3: Square Both Sides

Square both sides of the equation to eliminate the square root. This gives:\[ (\sqrt{x-7})^2 = \left(\frac{10 - x}{2}\right)^2 \]Simplifying, we get:\[ x - 7 = \frac{(10-x)^2}{4} \]

Step 4: Clear the Fraction

Multiply both sides of the equation by 4 to clear the fraction:\[ 4(x-7) = (10-x)^2 \]Simplify the left side:\[ 4x - 28 = (10-x)^2 \]

Step 5: Expand and Simplify

Expand the right side and simplify the equation:\[ 4x - 28 = 100 - 20x + x^2 \]

Step 6: Bring All Terms to One Side

Rearrange the equation to bring all terms to one side, forming a quadratic equation:\[ x^2 - 24x + 128 = 0 \]

Step 7: Solve the Quadratic Equation

Use the quadratic formula to solve for \(x\). The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -24\), and \(c = 128\).Calculate the discriminant: \(b^2 - 4ac = (-24)^2 - 4 \times 1 \times 128 = 576 - 512 = 64\). Since the discriminant is positive, there are two real solutions:\[ x = \frac{24 \pm \sqrt{64}}{2} \]\[ x = \frac{24 \pm 8}{2} \]Therefore, the solutions are \(x = 16\) and \(x = 8\).

Step 8: Verify the Solutions

Substitute \(x = 16\) and \(x = 8\) back into the original equation to ensure they are valid solutions.For \(x = 16\):\[ 16 + 2\sqrt{16-7} = 16 + 6 = 22 eq 10 \]For \(x = 8\):\[ 8 + 2\sqrt{8-7} = 8 + 2 = 10 \quad \text{(solution is valid)} \]Only \(x = 8\) is a valid solution.

Key Concepts

Quadratic Equation

Quadratic equations are a key concept in algebra. They take the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable. These equations might look intimidating at first, but they're quite manageable once you understand how to work with them.

• A quadratic equation is characterized by the highest power of the variable, which is 2.
• This type of equation can have zero, one, or two real solutions.
• Real solutions refer to the values of \(x\) that make the entire equation true.

Solving a quadratic equation involves techniques such as factoring, completing the square, or using the quadratic formula. The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), is a powerful tool when the equation does not factor easily. The term under the square root, \(b^2 - 4ac\), is called the discriminant and determines the nature of the solutions:

• If the discriminant is positive, there are two distinct real solutions.
• If zero, there's exactly one real solution.
• If negative, the solutions are not real numbers.

Understanding these properties of quadratic equations helps in solving them efficiently.

Square Root Isolation

Square root isolation is an essential step when solving equations involving radicals. The goal is to isolate the square root term on one side of the equation. This ensures you can eliminate the square root by squaring both sides in subsequent steps. Consider the equation \(x + 2 \sqrt{x-7} = 10\).

• Subtract \(x\) from both sides to isolate the square root term: \(2 \sqrt{x-7} = 10 - x\).
• Divide by 2 to further simplify: \(\sqrt{x-7} = \frac{10 - x}{2}\).

By isolating the square root, you simplify the problem and make it manageable to square both sides without changing the equation's balance. Once isolated, the next step is to remove the radical by squaring each side.

• This leads directly into forming a standard algebraic equation, which is often quadratic.
• Clearing a square root is crucial for solving equations that initially involve radicals.

Always remember to check the solutions afterward, as squaring both sides can sometimes introduce extraneous solutions not valid in the original equation.

Verification of Solutions

Verifying the solutions found from an equation is an important step, especially when dealing with squared terms. This process ensures that the solutions satisfy the original equation and are not extraneous. Let's use the solutions from the equation \(x + 2 \sqrt{x-7} = 10\):

• After solving, you found potential solutions: \(x = 16\) and \(x = 8\).
• Verification involves substituting these values back into the original equation.

For \(x = 16\): Check if \(16 + 2\sqrt{16-7} = 10\). Calculating gives \(16 + 6 = 22\), which does not equal 10. Hence, \(x = 16\) is not a valid solution.
For \(x = 8\): Check if \(8 + 2\sqrt{8-7} = 10\). Here, \(8 + 2 = 10\), which matches perfectly. Thus, \(x = 8\) is the valid solution.

• Always verify to identify any invalid or extraneous solutions introduced by operations like squaring.
• Ensuring the accuracy of solutions helps in confirming the correctness of your work.

Verification helps in reaffirming that the final solution set indeed solves the original equation without any errors.