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Chapter 11: Problem 70

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

Short Answer

Expert verified
The number that satisfies the given conditions is \(x = -1/2\).

Step by step solution

01

Frame the equation

Let's let the desired negative number be \(x\). Thus, the problem statement can be translated into the following equation: \(2x^2 - (1 + 2x) = 0\). Rearrange the equation, we have: \(2x^2 - 2x - 1 = 0\).
02

Apply the Quadratic Formula

The quadratic formula is \(x = [-b \pm \sqrt{b^2 -4ac}]/(2a)\), which we can apply to solve the equation \(2x^2 - 2x - 1 = 0\) (where \(a = 2\), \(b = -2\), and \(c = -1\)).
03

Solve the Equation

Substitute \(a\), \(b\), and \(c\) into the quadratic formula, we have two solutions: \(x_1 = [2 + \sqrt{(-2)^2 - 4*2*(-1)}]/(2*2) = 1\), \(x_2 = [2 - \sqrt{(-2)^2 - 4*2*(-1)}]/(2*2) = -1/2\). Notice: because we only consider negative numbers, we take \(x = -1/2\) as the final solution.

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