/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 130 Solve the equation using any con... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the equation using any convenient method. $$ x^{2}+3 x-\frac{3}{4}=0 $$

Short Answer

Expert verified
The solutions to the equation are \(x = 0.232\) and \(x = -3.232\).

Step by step solution

01

Identify a, b, and c from the given equation

From the quadratic equation \(x^{2}+3x-\frac{3}{4}=0\), we can identify that a is 1, b is 3, and c is -\frac{3}{4}.
02

Substitute a, b and c values into the quadratic formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). After substituting our values into this formula, we get \(x = \frac{-3 \pm \sqrt{(3)^2 - 4*(1)*-\frac{3}{4}}}{2*1}\).
03

Simplify the expression inside the square root

Simplify the expression within the square root. This gives us \(x = \frac{-3 \pm \sqrt{9 + 3}}{2}\).
04

Simplify further and solve for x

In simplifying further, we get \(x = \frac{-3 \pm \sqrt{12}}{2}\). Solving using the positive and negative roots, we get two possible solutions for x; \(x = \frac{-3 + \sqrt{12}}{2}\) and \(x = \frac{-3 - \sqrt{12}}{2}\). These solutions simplify to \(x = -1.5 + 1.732\) and \(x = -1.5 - 1.732\).

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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 vital tool for solving quadratic equations. Quadratic equations are in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. The quadratic formula is expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides the solution(s) to any quadratic equation by calculating the roots. The "+" and "-" signs indicate that the formula can yield two solutions, also known as the roots of the equation. You start by identifying the coefficients \( a \), \( b \), and \( c \) from the equation. Then substitute these values into the formula.
With this method, calculations become systematic and clear, reducing errors in finding the roots of the quadratic. The hardest part is often managing the arithmetic, especially under the square root. However, as you practice, using the formula will come easily and can solve complex-looking problems in no time.
Roots of a Quadratic
The roots of a quadratic equation are the values of \( x \) that make the equation true (i.e., the equation equals zero). When using the quadratic formula, once you've substituted the values of \( a \), \( b \), and \( c \) and simplified, two roots emerge.
These roots may be:
  • Real and distinct
  • Real and identical (when there is only one solution)
  • Complex (involving imaginary numbers when the discriminant is negative)
In the exercise provided, the roots are real and distinct, obtained from \( x = \frac{-3 \pm \sqrt{12}}{2} \). They approximate to \( x = -1.5 + 1.732 \) and \( x = -1.5 - 1.732 \). Calculating precisely can be tricky, so always double-check your arithmetic for accuracy.
Solving Quadratics
Solving quadratics involves finding the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \). There are several methods to solve these, including:
  • Factoring: For simple quadratics that can be rewritten into two binomials.
  • Completing the square: Transforming the quadratic into a perfect square trinomial.
  • Quadratic formula: A straightforward method used in our current exercise.
Here, we applied the quadratic formula for a quick and reliable solution. Factoring wasn't feasible because of the non-integer coefficients. Practicing various methods allows flexibility in approach and a deeper understanding of quadratic behavior.
Algebraic Manipulation
Algebraic manipulation is key to solving quadratic equations, especially when dealing with the quadratic formula. It involves:
  • Simplifying expressions: Ensure all terms are tidy and easy to work with.
  • Handling square roots: Carefully simplifying under the root to find exact or approximate answers.
  • Performing arithmetic: Accurately calculating each component of the quadratic formula.
In the example from the exercise, after identifying \( a \), \( b \), and \( c \), substituting correctly, and breaking down the expression under the root, each algebraic step brought us closer to the solution. It's crucial to take each step one at a time and verify your calculations. With practice, these manipulations become second nature, and you'll solve quadratics efficiently.

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Most popular questions from this chapter

Average Cost The inventor of a new game believes that the variable cost for producing the game is \(\$ 0.95\) per unit and the fixed costs are \(\$ 6000\). The inventor sells each game for \(\$ 1.69\). Let \(x\) be the number of games sold. (a) The total cost for a business is the sum of the variable cost and the fixed costs. Write the total cost \(C\) as a function of the number of games sold. (b) Write the average cost per unit \(\bar{C}=C / x\) as a function of \(x\).

(a) use the position equation \(s=-16 t^{2}+v_{0} t+s_{0}\) to write a function that represents the situation, (b) use a graphing utility to graph the function, (c) find the average rate of change of the function from \(t_{1}\) to \(t_{2}\), (d) interpret your answer to part (c) in the context of the problem, (e) find the equation of the secant line through \(t_{1}\) and \(t_{2}\), and (f) graph the secant line in the same viewing window as your position function. An object is thrown upward from a height of \(6.5\) feet at a velocity of 72 feet per second. $$ t_{1}=0, t_{2}=4 $$

In Exercises 55-68, determine whether the function has an inverse function. If it does, find the inverse function. $$ f(x)= \begin{cases}-x, & x \leq 0 \\ x^{2}-3 x, & x>0\end{cases} $$

Temperature The table shows the temperature \(y\) (in degrees Fahrenheit) of a certain city over a 24-hour period. Let \(x\) represent the time of day, where \(x=0\) corresponds to \(6 \mathrm{~A}\).M. $$ \begin{array}{|c|c|} \hline \text { Time, } \boldsymbol{x} & \text { Temperature, } \boldsymbol{y} \\\ \hline 0 & 34 \\ 2 & 50 \\ 4 & 60 \\ 6 & 64 \\ 8 & 63 \\ 10 & 59 \\ 12 & 53 \\ 14 & 46 \\ 16 & 40 \\ 18 & 36 \\ 20 & 34 \\ 22 & 37 \\ 24 & 45 \\ \hline \end{array} $$ A model that represents these data is given by \(y=0.026 x^{3}-1.03 x^{2}+10.2 x+34, \quad 0 \leq x \leq 24 .\) (a) Use a graphing utility to create a scatter plot of the data. Then graph the model in the same viewing window. (b) How well does the model fit the data? (c) Use the graph to approximate the times when the temperature was increasing and decreasing. (d) Use the graph to approximate the maximum and minimum temperatures during this 24 -hour period. (e) Could this model be used to predict the temperature for the city during the next 24 -hour period? Why or why not?

Use a graphing utility to graph each function. Write a paragraph describing any similarities and differences you observe among the graphs. (a) \(y=x\) (b) \(y=x^{2}\) (c) \(y=x^{3}\) (d) \(y=x^{4}\) (e) \(y=x^{5}\) (f) \(y=x^{6}\)

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