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Chapter 0: Problem 110

Use the Quadratic Formula to solve the equation. $$ 3 x+x^{2}-1=0 $$

Short Answer

Expert verified
The solutions to the equation \( x^2 + 3x - 1 = 0 \) are \( x1 = [-3 + sqrt(13)] / 2 \) and \( x2 = [-3 - sqrt(13)] / 2 \)

Step by step solution

01

Identify the coefficients

The equation \( x^2 + 3x - 1 = 0 \) has the coefficients \( a = 1 \), \( b = 3 \), and \( c = -1 \).
02

Apply the quadratic formula

The quadratic formula is \( x = [-b 卤 sqrt(b^2 - 4ac)] / (2a) \). Substituting the known values of the coefficients, \( x = [-3 卤 sqrt((3)^2 - 4*1*(-1))] / (2*1) = [-3 卤 sqrt(9 + 4)] / 2 = [-3 卤 sqrt(13)] / 2.
03

Simplify the results

The two potential values for x are \( x1 = [-3 + sqrt(13)] / 2 \) and \( x2 = [-3 - sqrt(13)] / 2 \). These are the solutions to the given quadratic equation.

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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 type of equation that involves a variable squared. Its standard form is written as \( ax^2 + bx + c = 0 \). Here, \( x \) represents the variable, and \( a \), \( b \), and \( c \) are constants. This kind of equation forms a parabola when graphed.

Quadratic equations can have zero, one, or two solutions, depending on the value of the discriminant. The discriminant is the part of the quadratic formula under the square root, \( b^2 - 4ac \).

Understanding the structure of a quadratic equation is key to solving it. By knowing how to work with these equations, you can find the points where the equation equals zero.
Coefficients Identification
Identifying the coefficients in a quadratic equation is an important first step in solving the equation. These are the numbers that multiply the terms of the variable in the equation. In the equation \( ax^2 + bx + c = 0 \), \( a \), \( b \), and \( c \) are the coefficients.

For example, in the equation \( x^2 + 3x - 1 = 0 \):
  • \( a = 1 \) is the coefficient of \( x^2 \).
  • \( b = 3 \) is the coefficient of \( x \).
  • \( c = -1 \) is the constant term.
Accurate identification of these coefficients is crucial because they are used in the quadratic formula to find the solutions for \( x \). Misidentifying these values can lead to incorrect solutions.
Solving Quadratic Equations
Once the coefficients are identified, you can apply the quadratic formula, which is a reliable method for finding the roots of any quadratic equation. The formula is:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]

By substituting \( a \), \( b \), and \( c \) into the formula, you can compute the value of \( x \). This formula accounts for both potential solutions, due to the \( 卤 \) symbol which signifies that you will calculate two different values:
  • \( x_1 = \frac{{-b + \sqrt{{b^2 - 4ac}}}}{2a} \)
  • \( x_2 = \frac{{-b - \sqrt{{b^2 - 4ac}}}}{2a} \)
Make sure to handle the arithmetic carefully. This will lead you directly to the solutions of the quadratic equation.
Simplifying Square Roots
Simplifying square roots, especially when using the quadratic formula, is an essential step to find clean and usable solutions. In the quadratic formula:

\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)

The expression \( b^2 - 4ac \) under the square root may not always be a perfect square. Here's what to do:
  • If the square root simplifies to a whole number, calculate it, and substitute the value back into the formula for crisp solutions.
  • If not, maintain it in root form, as seen in solutions like \( \frac{{-3 \pm \sqrt{13}}}{2} \).
Working through these roots carefully helps ensure that your final answers are accurate and as simplified as possible.

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Most popular questions from this chapter

The set of ordered pairs \(\\{(-8,-2),(-6,0),(-4,0)\), \((-2,2),(0,4),(2,-2)\\}\) represents a function.

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=x^{5}-2 $$

In Exercises 69-74, use the functions given by \(f(x)=\frac{1}{8} x-3\) and \(g(x)=x^{3}\) to find the indicated value or function. $$ g^{-1} \circ f^{-1} $$

Find the average rate of change of the function from \(x_{1}\) to \(x_{2}\). $$ \begin{array}{cc} \text { Function } & x \text {-Values } \\ f(x)=x^{3}-3 x^{2}-x &\quad x_{1}=1, x_{2}=3 \end{array} $$

In Exercises 39-54, (a) find the inverse function of \(f\), (b) graph both \(f\) and \(f^{-1}\) on the same set of coordinate axes, (c) describe the relationship between the graphs of \(f\) and \(f^{-1}\), and (d) state the domain and range of \(f\) and \(f^{-1}\). $$ f(x)=\frac{4}{x} $$
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