Problem 126 - College Algebra

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 solution to the equation is \(x = 0.5 - 0.5\sqrt{3}\).

Step 1: Rewriting the problem in equation format

We know the problem states 'When the sum of 1 and twice a negative number is subtracted from twice the square of the number, 0 results'. This can be expressed as '2x^2 - (2x+1) = 0'.

Step 2: Simplifying the equation

When we simplify the equation, it should simplify into the standard form of a quadratic equation, which is \[ax^2 + bx + c = 0\]. So, our equation simplifies to \(2x^2 - 2x - 1 = 0\).

Step 3: Apply the Quadratic Formula

Applying the Quadratic Formula to our equation \(2x^2 - 2x - 1 = 0\), where a = 2, b = -2, and c = -1, gives us \[x=\frac{-(-2) \pm \sqrt{(-2)^2 - 4*2*-1}}{2*2}\]. This further simplifies to \[x = \frac{2 \pm \sqrt{4+8}}{4}\]

Step 4: Solve for x

Solving for x gives us \[x = \frac{2 \pm \sqrt{12}}{4}\]. This can be further simplified to \[x = \frac{2 \pm 2\sqrt{3}}{4}\] which simplifies further to \[x = 0.5 \pm 0.5\sqrt{3}\]. As the problem mentions a negative number, the only solution we consider is \[x = 0.5 - 0.5\sqrt{3}\].