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Question 7 of 10, Step 2 of 2 Consider the following quadratic equation: Step 2 of \( 2:-4 x^{2}+2=x \) formula \( x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \)

Ask by Harrington Fernandez. in the United States
Mar 17,2025

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The solutions are \( x = -\frac{1 + \sqrt{33}}{8} \) and \( x = -\frac{1 - \sqrt{33}}{8} \).

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We start with the equation \[ -4 x^2 + 2 = x. \] First, rearrange the equation so that all terms are on one side: \[ -4 x^2 - x + 2 = 0. \] Here, we identify the coefficients: \[ a = -4,\quad b = -1,\quad c = 2. \] Next, substitute these into the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Substitute \( a = -4 \), \( b = -1 \), and \( c = 2 \): \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(-4)(2)}}{2(-4)}. \] Simplify the numerator: \[ -(-1) = 1. \] Calculate the discriminant: \[ (-1)^2 - 4(-4)(2) = 1 + 32 = 33. \] Now, the formula becomes: \[ x = \frac{1 \pm \sqrt{33}}{-8}. \] This can also be written by factoring out a \(-1\) in the numerator: \[ x = -\frac{1 \pm \sqrt{33}}{8}. \] Thus, the solutions are: \[ x = -\frac{1 + \sqrt{33}}{8} \quad \text{or} \quad x = -\frac{1 - \sqrt{33}}{8}. \]

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Did you know that quadratic equations have been studied for thousands of years? Ancient civilizations like the Babylonians were solving problems that eventually would be modeled by quadratic equations, albeit using geometrical methods rather than algebraic formulas. This shows that our approach to solving these equations has deep historical roots! In real-world applications, quadratic equations pop up everywhere—from calculating the trajectory of projectiles to modeling population growth and profit maximization in businesses. Anytime you have a situation where the relationship between variables forms a curved graph, you can bet there's likely a quadratic equation at play, making them invaluable tools in practical scenarios!

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