Problem 126 - Algebra and Trigonometry

Question

When the sum of 1 and twice a negative number is subtracted from twice the square of the number, 0 results. Find the number.

Step-by-Step Solution

Answer

The negative number is \(-0.5 - 0.5 \sqrt{3}\) or \(-0.5 + 0.5 \sqrt{3}\), because as a negative number it should be less than 0.

Step 1: Express the problem as an equation

The problem can be written as follows: 2x^2 - (1 + 2x) = 0. 2x^2 represents twice the square of the negative number. -(1 + 2x) represents the subtraction of the sum of 1 and twice the negative number from twice the square of the number.

Step 2: Simplify the equation

Combine like terms as follows: 2x^2 - 1 - 2x = 0, which can be re-arranged as 2x^2 - 2x - 1= 0.

Step 3: Solve the quadratic equation

The quadratic equation can be solved by using the quadratic formula, given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plugging a = 2, b = -2, and c = -1 into the formula, we have \(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 2 \times (-1)}}{2 \times 2} = \frac{2 \pm \sqrt{4 + 8}}{4} = \frac{2 \pm \sqrt{12}}{4} = \frac{2 \pm 2 \sqrt{3}}{4}\), which simplifies to \(x = 0.5 \pm 0.5 \sqrt{3}\).