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Chapter 2: Problem 38

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution. $$ 2 x^{2}+5 x+3=0 $$

Short Answer

Expert verified
The solutions are \( x = -1 \) and \( x = -\frac{3}{2} \).

Step by step solution

01

Identify the Quadratic Equation

The quadratic equation given is \( 2x^2 + 5x + 3 = 0 \). This equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 2 \), \( b = 5 \), and \( c = 3 \).
02

Write the Quadratic Formula

The quadratic formula is used to solve equations of the form \( ax^2 + bx + c = 0 \). It is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
03

Calculate the Discriminant

The discriminant of a quadratic equation, \( b^2 - 4ac \), helps determine the nature of the roots. For the equation \( 2x^2 + 5x + 3 = 0 \), calculate the discriminant:\[ b^2 - 4ac = 5^2 - 4 \times 2 \times 3 = 25 - 24 = 1 \]
04

Determine the Nature of the Roots

Since the discriminant is \( 1 \), which is greater than zero, the equation has two distinct real roots.
05

Apply the Quadratic Formula

Substitute \( a = 2 \), \( b = 5 \), and \( c = 3 \) into the quadratic formula:\[ x = \frac{-5 \pm \sqrt{1}}{4} \]
06

Calculate the Roots

Evaluate the expression for the roots:- First root: \( x = \frac{-5 + 1}{4} = \frac{-4}{4} = -1 \)- Second root: \( x = \frac{-5 - 1}{4} = \frac{-6}{4} = -\frac{3}{2} \)

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Key Concepts

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

Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable with the form \( ax^2 + bx + c = 0 \). It features three key components: the quadratic term \( ax^2 \), where \( a eq 0 \); the linear term \( bx \); and the constant term \( c \). The graph of a quadratic equation is a parabola that can open upwards or downwards depending on the sign of coefficient \( a \).
  • If \( a > 0 \), the parabola opens upwards.
  • If \( a < 0 \), it opens downwards.
Quadratic equations are foundational in algebra and have applications across various fields, from physics to economics. They can be solved through several methods, including factoring, completing the square, and using the quadratic formula.
Discriminant
The discriminant is a key part of the quadratic formula that helps determine the nature of the roots of a quadratic equation. It is given by the expression \( b^2 - 4ac \), where \( a \), \( b \), and \( c \) are the coefficients of the equation \( ax^2 + bx + c = 0 \). The discriminant provides insight into the type and number of roots the equation has:
  • If \( b^2 - 4ac > 0 \), there are two distinct real roots.
  • If \( b^2 - 4ac = 0 \), there is exactly one real root, which implies a repeated, or double root.
  • If \( b^2 - 4ac < 0 \), the roots are complex and not real.
By computing the discriminant, you can predict the nature of the solutions without actually solving the equation.
Real Roots
Real roots are solutions to a quadratic equation where the values satisfy the equation when substituted back. They are found using the quadratic formula or other methods only when the discriminant \( b^2 - 4ac \) is zero or positive.
In our example, the discriminant is calculated to be \( 1 \), which is greater than zero, indicating the presence of two distinct real roots.
To extract these roots from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), you solve:
  • The root \( x_1 \) derived from the \( + \) sign.
  • The root \( x_2 \) derived from the \( - \) sign.
Real roots are significant as they signify actual points where the graph of the quadratic intersects the x-axis.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). This form is pivotal because it sets the stage for various solution methods, including the quadratic formula.
To effectively use the quadratic formula, a quadratic equation must be written in this standard form, where
  • \( a \) is the coefficient of \( x^2 \), known as the leading coefficient, and cannot be zero.
  • \( b \) is the coefficient of \( x \), the linear term.
  • \( c \) is the constant term.
Recognizing and transforming equations into standard form enable seamless application of algebraic techniques and tools like the quadratic formula to find solutions.

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Most popular questions from this chapter

For the following exercises, solve the compound inequality. Express your answer using inequality signs, and then write your answer using interval notation. $$ 3 x+1>2 x-5>x-7 $$

For the following exercises, solve the inequality. Write your final answer in interval notation. $$ \frac{x-1}{3}+\frac{x+2}{5} \leq \frac{3}{5} $$

For the following exercises, solve the following polynomial equations by grouping and factoring. $$ 3 x^{3}-6 x^{2}-27 x+54=0 $$

For the following exercises, graph both straight lines (left-hand side being y1 and right-hand side being \(y 2\) ) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the \(y\) -values of the lines. $$ x+3<3 x-4 $$

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. $$ \left(x^{2}-1\right)^{2}+\left(x^{2}-1\right)-12=0 $$
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