Problem 34 - Algebra and Trigonometry

Question

\(5-60\) 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 term

Start by isolating the square root term on one side of the equation. Subtract \( x \) from both sides to get: \[2\sqrt{x-7} = 10 - x\]

Step 2: Remove the square root

Square both sides of the equation to eliminate the square root, yielding:\[4(x-7) = (10 - x)^2\] This expands to:\[ 4x - 28 = 100 - 20x + x^2 \]

Step 3: Rearrange the quadratic equation

Bring all terms to one side to form a standard quadratic equation:\[ x^2 - 24x + 128 = 0 \]

Step 4: Solve the quadratic equation by the quadratic formula

Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -24 \), and \( c = 128 \).Calculate the discriminant:\[(-24)^2 - 4 \cdot 1 \cdot 128 = 576 - 512 = 64\]Thus, the solutions are:\[x = \frac{24 \pm \sqrt{64}}{2} = \frac{24 \pm 8}{2} \]

Step 5: Find exact solutions from the quadratic formula

Calculate both potential solutions:\[x = \frac{24 + 8}{2} = 16\]\[x = \frac{24 - 8}{2} = 8\]

Step 6: Verify solutions

Substitute each solution back into the original equation to verify it:For \( x = 16 \):\[16 + 2\sqrt{16 - 7} = 16 + 2\sqrt{9} = 16 + 6 = 22 \quad (not \: equal \: to \: 10)\]For \( x = 8 \):\[8 + 2\sqrt{8 - 7} = 8 + 2 \cdot 1 = 10 \, \text{(valid solution)}\]

Key Concepts

Quadratic Formula

The Quadratic Formula is a powerful tool used to solve quadratic equations. These equations take the standard form \( ax^2 + bx + c = 0 \). To find the solutions, substitute values for \( a \), \( b \), and \( c \) into the formula:

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

This formula helps find both real and complex solutions. The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It is crucial to determining the nature of the solutions.
For example, in our exercise, the quadratic equation derived was \( x^2 - 24x + 128 = 0 \). Here, \( a = 1 \), \( b = -24 \), and \( c = 128 \). The discriminant calculated was \( 64 \), which is positive, indicating two real solutions. By plugging these values into the formula, we obtained the solutions \( x = 16 \) and \( x = 8 \). Only the solution \( x = 8 \) satisfied the original equation.

Isolating Square Roots

Isolating the square root is an essential step when dealing with equations that contain a square root term. To do this, you need to simplify the equation so that the square root stands alone on one side.
Consider the equation \( x + 2\sqrt{x-7} = 10 \). To isolate the square root, subtract \( x \) from both sides, giving:

• \( 2\sqrt{x-7} = 10 - x \)

This step simplifies the handling of the equation by focusing on removing the square root through squaring both sides.
Once the square root term is isolated, we can then eliminate it by squaring the entire equation, which allowed us to derive the quadratic equation in our example.

Real Solutions

In algebra, real solutions are those solutions that do not involve imaginary components. To determine if solutions are real, we look at the discriminant of the quadratic equation:

• If the discriminant \( b^2 - 4ac \) is positive, the equation has two distinct real solutions.
• If it equals zero, there is exactly one real solution.
• If negative, the solutions are not real but complex or imaginary.

For the equation \( x^2 - 24x + 128 = 0 \), the discriminant was \( 64 \), thus predicting two real solutions. We calculated these solutions, \( x = 8 \) and \( x = 16 \), using the quadratic formula. However, only \( x = 8 \) was valid when substituted back into the original equation, confirming it as the real solution.