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Chapter 1: Problem 68

In Exercises \(67-69\), solve each equation. \(2 x^{2}+x=6\) (Section P.7, Example 7)

Short Answer

Expert verified
The solution to the equation \(2x^{2} + x - 6 = 0\) are \(x = 1.5, -2.\)

Step by step solution

01

Rearrange the given equation

To solve the quadratic equation, we must first transfer it into its standard form. Which gives: \(2x^{2} + x - 6 = 0\)
02

Identify values of a, b and c

From the standard form of the quadratic equation, we can identify \(a = 2\), \(b = 1\), and \(c = -6\).
03

Calculate the determinant

The determinant, also known as discriminant is given as \(\Delta = b^{2} - 4ac\). So, calculating it will give us: \(\Delta = (1)^{2} - 4 * 2 * -6 = 1 + 48 = 49\).
04

Apply the quadratic formula

The solutions for \(x\) can be found using the quadratic formula \(-b \pm \sqrt{\Delta}/(2a)\). As \(\Delta > 0\), there will be two distinct real roots. So we get: \(x = [-1+\sqrt{49}]/(2*2) = 1.5, [-1-\sqrt{49}]/(2*2) = -2\)

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Key Concepts

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

Discriminant
When solving a quadratic equation, the discriminant helps you understand the nature of the roots without actually having to find them first. For the quadratic equation in standard form, \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) is determined by the formula \(\Delta = b^2 - 4ac\).
\[\text{The discriminant helps to determine several root characteristics:}\]
  • If \(\Delta > 0\), the quadratic equation has two distinct real roots.
  • If \(\Delta = 0\), the equation has exactly one real root, or you might hear it as a repeated root.
  • If \(\Delta < 0\), the equation has no real roots; instead, it has two complex (imaginary) roots.
Understanding the discriminant is a great way to predict the nature of solutions without actually solving the equation. In the solved equation from the exercise, the discriminant is 49, which is greater than zero, indicating two distinct real solutions.
Quadratic Formula
The quadratic formula is a universal tool for solving quadratic equations, especially when factoring seems complicated or impossible. It is expressed as:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula stems directly from rearranging a quadratic equation into its standard form and then using algebraic manipulation.

Steps to use the quadratic formula
  • First, ensure the equation is in the standard form \(ax^2 + bx + c = 0\).
  • Identify the coefficients: \(a\), \(b\), and \(c\).
  • Calculate the discriminant \(\Delta = b^2 - 4ac\).
  • Plug these values into the quadratic formula.
  • Solve for \(x\) taking both the positive and negative values of the square root, giving you two potential solutions.
In our exercise, satisfying the quadratic formula provided solutions \(x = 1.5\) and \(x = -2\), as accurately calculated by substituting the values into the formula.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is a self-contained format that looks like this: \(ax^2 + bx + c = 0\).
Here, \(a\), \(b\), and \(c\) are constants with \(a eq 0\) to ensure the equation remains quadratic.

Placing a quadratic in standard form is the crucial first step needed to effectively solve it, whether through factoring, completing the square, or applying the quadratic formula.

Key Purposes of Standard Form:
  • Identifying coefficients: Helps in easily picking out \(a\), \(b\), and \(c\) which are essential in further calculations like using the quadratic formula.
  • Simplifying the solving process: Once in standard form, a variety of solving methods become applicable.
  • Analyzing graph behavior: It makes it easier to determine the upward or downward opening of the parabola, as well as the vertical intercept \(y = c\).
In the provided exercise, we first rearranged \(2x^2 + x = 6\) into \(2x^2 + x - 6 = 0\) to conform with the standard form. This allowed us to proceed with the solution using the defined coefficients.

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Most popular questions from this chapter

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