Chapter 7: Problem 16
solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$3 x^{2}=3-4 x$$
Short Answer
Expert verified
The solutions are \( x = \frac{-2 + \sqrt{13}}{3} \) and \( x = \frac{-2 - \sqrt{13}}{3} \).
Step by step solution
01
Rearrange the Equation
First, we need to rearrange the given equation into the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). The given equation is \( 3x^2 = 3 - 4x \). Let's move all terms to one side to get: \( 3x^2 + 4x - 3 = 0 \).
02
Identify Coefficients
Identify the coefficients from the standard form of the quadratic equation \( 3x^2 + 4x - 3 = 0 \). Here, \( a = 3 \), \( b = 4 \), and \( c = -3 \).
03
Apply the Quadratic Formula
The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 3 \), \( b = 4 \), and \( c = -3 \) into this formula.
04
Solve the Discriminant
Calculate the discriminant using the formula \( b^2 - 4ac \). Here, we have \( 4^2 - 4 \times 3 \times (-3) = 16 + 36 = 52 \).
05
Compute the Roots
Now substitute the discriminant and other values into the quadratic formula: \[ x = \frac{-4 \pm \sqrt{52}}{6} \]. Simplify \( \sqrt{52} \) to \( 2\sqrt{13} \), resulting in two solutions: \[ x = \frac{-4 + 2\sqrt{13}}{6} \] and \[ x = \frac{-4 - 2\sqrt{13}}{6} \].
06
Simplify the Expression
Further simplify the expressions for \( x \): \[ x = \frac{-2 + \sqrt{13}}{3} \] and \[ x = \frac{-2 - \sqrt{13}}{3} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations are a central concept in algebra. These are polynomial equations of degree two, meaning they have a variable raised to the power of two. Typically, they are written in the standard form:
For example, the equation \( 3x^2 = 3 - 4x \) can be rearranged into standard form as:
- \( ax^2 + bx + c = 0 \)
- Here, \( a \), \( b \), and \( c \) are constants.
For example, the equation \( 3x^2 = 3 - 4x \) can be rearranged into standard form as:
- \( 3x^2 + 4x - 3 = 0 \)
Discriminant Calculation
The discriminant is a key part of understanding the nature of the solutions to a quadratic equation. It is found within the quadratic formula itself:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- Here, \( b^2 - 4ac \) is the discriminant.
- If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
- If \( b^2 - 4ac = 0 \), there is one real solution.
- If \( b^2 - 4ac < 0 \), there are no real solutions (the solutions are complex).
- \( b^2 - 4ac = 16 + 36 = 52 \)
Solving Quadratics
Solving quadratic equations using the quadratic formula is a reliable method that works for any quadratic equation. The formula is:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( a = 3 \), \( b = 4 \), \( c = -3 \)
- Substitute these into the formula:
- \( x = \frac{-4 \pm \sqrt{52}}{6} \)
- Simplifying further gives:
- \( x = \frac{-2 + \sqrt{13}}{3} \) and \( x = \frac{-2 - \sqrt{13}}{3} \)
Mathematics Education
Understanding quadratic equations is a fundamental part of mathematics education. It helps build a foundation for more advanced areas of study, such as calculus and linear algebra.
Educators often emphasize the different methods for solving quadratics to ensure that students have a versatile set of tools.
Educators often emphasize the different methods for solving quadratics to ensure that students have a versatile set of tools.
- Factoring: Involves expressing the equation as a product of its factors.
- Completing the Square: Rewrites the equation in a form that makes finding the roots more straightforward.
- Quadratic Formula: A universal method applicable to all quadratic equations.