Chapter 0: Problem 5
Use the quadratic formula to find the zeros of the functions in Exercises. $$ f(x)=-2 x^{2}+3 x-4 $$
Short Answer
Expert verified
The zeros are \( x = \frac{3}{4} - \frac{i\sqrt{23}}{4} \) and \( x = \frac{3}{4} + \frac{i\sqrt{23}}{4} \).
Step by step solution
01
Identify coefficients
For the quadratic function \(f(x) = -2x^2 + 3x - 4\), identify the coefficients: \(a = -2\), \(b = 3\), and \(c = -4\).
02
Write down the quadratic formula
The quadratic formula is used to find the solutions (or zeros) of a quadratic equation \(ax^2 + bx + c = 0\). The formula is given by: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
03
Substitute the coefficients into the formula
Substitute \(a = -2\), \(b = 3\), and \(c = -4\) into the quadratic formula: \[ x = \frac{-(3) \pm \sqrt{(3)^2 - 4(-2)(-4)}}{2(-2)} \]
04
Simplify inside the square root
Perform the operations inside the square root: \[ x = \frac{-3 \pm \sqrt{9 - 32}}{-4} = \frac{-3 \pm \sqrt{-23}}{-4} \]
05
Handle complex solutions
Since \(-23\) is negative, the square root will be an imaginary number. Therefore, \( \sqrt{-23} = i\sqrt{23} \): \[ x = \frac{-3 \pm i\sqrt{23}}{-4} \]
06
Split into two separate solutions
Split the equation into its two possible solutions: \[ x = \frac{-3 + i\sqrt{23}}{-4} \text{ and } x = \frac{-3 - i\sqrt{23}}{-4} \]
07
Simplify the solutions
Simplify each solution if possible: \[ x = \frac{3}{4} - \frac{i\sqrt{23}}{4} \text{ and } x = \frac{3}{4} + \frac{i\sqrt{23}}{4} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
coefficients
In the context of a quadratic equation, a coefficient is a numerical or constant value placed in front of variables like x. When you look at a quadratic equation in the form of\(ax^2 + bx + c = 0\), the letters a, b, and c are coefficients.
Each coefficient has a specific role:
To solve a quadratic equation using the quadratic formula, you need to correctly identify these coefficients. For example, given the function \(f(x) = -2x^2 + 3x - 4\), the coefficients are:
Each coefficient has a specific role:
- a: The leading coefficient multiplies \(x^2\).
- b: This coefficient multiplies x.
- c: The constant term, which stands alone.
To solve a quadratic equation using the quadratic formula, you need to correctly identify these coefficients. For example, given the function \(f(x) = -2x^2 + 3x - 4\), the coefficients are:
- a = -2
- b = 3
- c = -4
complex solutions
Complex solutions arise when the discriminant in the quadratic formula is negative. The discriminant is the part of the quadratic formula under the square root: \(b^2 - 4ac\).
If this value is negative, the equation has complex (or imaginary) solutions because the square root of a negative number is not a real number. Instead, it involves the imaginary unit \(i\), where \(i = \sqrt{-1}\).
For example, in our solved exercise:
If this value is negative, the equation has complex (or imaginary) solutions because the square root of a negative number is not a real number. Instead, it involves the imaginary unit \(i\), where \(i = \sqrt{-1}\).
For example, in our solved exercise:
- The discriminant is calculated as \(3^2 - 4(-2)(-4) = 9 - 32 = -23\).
- Since -23 is negative, the solutions will be complex.
- We write this as \(\sqrt{-23} = i\sqrt{23}\).
- \(\frac{3}{4} - \frac{i\sqrt{23}}{4}\)
- \(\frac{3}{4} + \frac{i\sqrt{23}}{4}\)
quadratic equation
A quadratic equation is a second-order polynomial equation in a single variable x, with the general form \(ax^2 + bx + c = 0\). Quadratic equations are fundamental in algebra and are widely used to model various real-world phenomena.
The solutions or 'roots' of a quadratic equation can be found using several methods, one of which is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides the solutions by taking into account the coefficients a, b, and c.
Here is the process step-by-step using our example \(f(x) = -2x^2 + 3x - 4\):
The solutions or 'roots' of a quadratic equation can be found using several methods, one of which is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula provides the solutions by taking into account the coefficients a, b, and c.
Here is the process step-by-step using our example \(f(x) = -2x^2 + 3x - 4\):
- Identify the coefficients: \(a = -2\), \(b = 3\), \(c = -4\).
- Plug these into the quadratic formula: \( x = \frac{-(3) \pm \sqrt{(3)^2 - 4(-2)(-4)}}{2(-2)} \).
- Simplify under the square root: \( x = \frac{-3 \pm \sqrt{9 - 32}}{-4} = \frac{-3 \pm \sqrt{-23}}{-4} \).