Problem 125 - Algebra and Trigonometry

Question

When the sum of 6 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

Step-by-Step Solution

Answer

The unknown positive number is 3.

Step 1: Translate the problem into equation

Let the positive number be denoted by x. Then according to the problem, the equation can be formulated as: \(x^2 - (6 + 2x) = 0\).

Step 2: Simplify the Equation

Simplify the equation by getting rid of the brackets and combining like terms, we have: \(x^2 - 6 - 2x = 0 \). After rearranging, we have \(x^2 - 2x - 6 = 0\).

Step 3: Solve the Quadratic Equation

Now, we can solve this quadratic equation using the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. We know that in the equation \(ax^2 + bx + c = 0\), a = 1, b = -2, and c = -6. After substituting these values, we have two solutions: \(x = \frac{2 \pm \sqrt{(-2)^2 - 4*1*(-6)}}{2*1}\). This simplifies to \(x = 3, -2\).

Step 4: Determine the Valid Solution

Since we are looking for a positive number, we reject -2. Therefore, the positive number is 3.