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Chapter 8: Problem 64

Solve. Approximate the solutions to three decimal places. \(x^{2}-0.75 x-0.5=0\)

Short Answer

Expert verified
The solutions are approximately 1.175 and -0.425.

Step by step solution

01

- Identify the quadratic equation

The given equation is in the form of a quadratic equation: \[ x^2 - 0.75x - 0.5 = 0 \]
02

- Use the quadratic formula

To find the solutions for the quadratic equation, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -0.75 \), and \( c = -0.5 \).
03

- Calculate the discriminant

The discriminant is calculated as: \[ b^2 - 4ac = (-0.75)^2 - 4(1)(-0.5) = 0.5625 + 2 = 2.5625 \]
04

- Calculate the solutions

Substitute the values into the quadratic formula: \[ x = \frac{-(-0.75) \pm \sqrt{2.5625}}{2(1)} = \frac{0.75 \pm 1.6}{2} \] This gives two solutions: \[ x_1 = \frac{0.75 + 1.6}{2} = 1.175 \] and \[ x_2 = \frac{0.75 - 1.6}{2} = -0.425 \]
05

- Approximate the solutions

The solutions to the equation \( x^2 - 0.75x - 0.5 = 0 \) approximately are: \[ x_1 \approx 1.175 \] and \[ x_2 \approx -0.425 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

quadratic formula
The quadratic formula is a useful tool for solving quadratic equations. A quadratic equation is of the form
\[ ax^2 + bx + c = 0 \]
where \( a, b, \) and \( c \) are constants. The quadratic formula provides the solution to this equation and is written as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula gives the values of \( x \) that make the quadratic equation true. The symbols \( \pm \) indicate that there will be two solutions: one with addition and one with subtraction.
Applying this formula is straightforward:
  • Identify coefficients \( a, b, \) and \( c \) from the equation

  • Substitute them into the formula

  • Simplify to find the values of \( x \)
In our problem, we have:
  • \( a = 1 \)

  • \( b = -0.75 \)

  • \( c = -0.5 \)
discriminant
The discriminant is a part of the quadratic formula that determines the nature of the roots of the equation. It is the expression under the square root in the quadratic formula:
\[ b^2 - 4ac \]
The value of the discriminant reveals important information:
  • If the discriminant is positive, there are two distinct real solutions.

  • If the discriminant is zero, there is exactly one real solution.

  • If the discriminant is negative, there are no real solutions; instead, the solutions are complex or imaginary.
In our example, we computed the discriminant as:
\[ (-0.75)^2 - 4(1)(-0.5) = 0.5625 + 2 = 2.5625 \]
Since 2.5625 is positive, the quadratic equation \( x^2 - 0.75x - 0.5 = 0 \) has two distinct real solutions.
solving quadratic equations
Solving quadratic equations involves finding their roots, which are the values of \( x \) that make the equation true. There are several methods to solve quadratic equations, including:
  • Factoring

  • Completing the square

  • Using the quadratic formula
For our problem, we used the quadratic formula. Here's a recap of the process:
  • Write the equation in the form \( ax^2 + bx + c = 0 \).

  • Identify coefficients \( a, b, \) and \( c \).

  • Calculate the discriminant \( b^2 - 4ac \).

  • Substitute \( a, b, \) and \( c \) into the quadratic formula.

  • Simplify all terms to find the solutions for \( x \).
Specifically,
\[ x = \frac{-(-0.75) \pm \sqrt{2.5625}}{2(1)} = \frac{0.75 \pm 1.6}{2} \]
We then obtained the approximate solutions:
\[ x_1 \approx 1.175 \]
\[ x_2 \approx -0.425 \]
These values of \( x \) satisfy the original quadratic equation.

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