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Chapter 11: Problem 69

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

Short Answer

Expert verified
The positive number that satisfies the given problem is \( 1 + 3\sqrt{7} \)

Step by step solution

01

Translate the Word Problem Into an Equation

Let's denote the positive number as \( x \). According to the problem, 'the sum of 6 and twice a positive number is subtracted from the square of the number', we interpret this as the equation; \( x^2 - (2x + 6) = 0 \)
02

Simplify the Equation

Simplify the equation to its standard form \( ax^2 + bx + c = 0 \). So, the equation becomes \( x^2 - 2x - 6 = 0 \)
03

Solve The Quadratic Equation

Use the quadratic formula \( x = (-b 卤 \sqrt{b^2 - 4ac}) / (2a) \) to solve the equation. Here a = 1, b = -2, and c = -6. Substituting these values into the formula, we get two solutions; \( x = (2 卤 \sqrt{(-2)^2 - 4 * 1 * -6}) / (2 * 1) \). Simplifying this further gives \( x = (2 卤 \sqrt{4 + 24}) / 2 = (2 卤 \sqrt{28}) / 2 \). Thus, we obtain two solutions \( x = 1 卤 3\sqrt{7} \). Since x is a positive number, we reject the negative solution.
04

Final Check

Upon obtaining the positive solution \( x = 1 + 3\sqrt{7} \), we need to check if it satisfies the original equation. Substituting \( x \) into the equation \( x^2 - 2x - 6 = 0 \), we see that both sides of the equation are equal, confirming the solution is correct.

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