/*! 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 21 For Activities 19 through \(26,\... [FREE SOLUTION] | 91Ó°ÊÓ

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For Activities 19 through \(26,\) solve for the input that corresponds to each of the given output values. (Round answers to three decimal places when appropriate.) $$ s(t)=\frac{10}{t+5}+6 t ; s(t)=18, s(t)=0 $$

Short Answer

Expert verified
The solution for \( s(t)=18 \) has no real roots. For \( s(t)=0 \), the solutions are real numbers found using the quadratic formula.

Step by step solution

01

Set up Equation for First Output

The first given output is 18. We need to solve the equation \( s(t) = 18 \). Using the function \( s(t) = \frac{10}{t+5} + 6t \), set this equal to 18:\[ \frac{10}{t+5} + 6t = 18 \]
02

Solve for t in the First Equation

Rearrange the equation from Step 1:\[ \frac{10}{t+5} = 18 - 6t \]Multiply through by \( t+5 \) to eliminate the fraction:\[ 10 = (18 - 6t)(t+5) \]Expand and simplify:\[ 10 = 18t + 90 - 6t^2 - 30t \]\[ 6t^2 + 12t + 80 = 0 \]Solve the quadratic equation using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6 \), \( b = 12 \), \( c = 80 \).
03

Solve Second Equation for Second Output

Now solve for the second given output 0 in the equation:\[ \frac{10}{t+5} + 6t = 0 \]Rearrange and eliminate the fraction:\[ \frac{10}{t+5} = -6t \]\[ 10 = -6t(t+5) \]\[ 10 = -6t^2 - 30t \]\[ 6t^2 + 30t + 10 = 0 \]Solve this quadratic equation similarly using the quadratic formula with \( a = 6 \), \( b = 30 \), and \( c = 10 \).
04

Calculate and Simplify Solutions

For both equations from Steps 2 and 3, apply the quadratic formula.* For the equation from Step 2: - Discriminant: \( 12^2 - 4 \cdot 6 \cdot 80 = 144 - 1920 \) - Calculate the roots (if the discriminant is negative, there are no real solutions).* For the equation from Step 3: - Discriminant: \( 30^2 - 4 \cdot 6 \cdot 10 = 900 - 240 \) - Calculate the two possible solutions for \( t \), simplifying each root.

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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 powerful tool used to determine the solutions of quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula comes in handy for solving equations that are not easily factorable. It allows you to find the roots of any quadratic equation by using the coefficients \( a \), \( b \), and \( c \), which are the numbers in front of the variables in the equation.

The general quadratic formula is given by:
  • \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Applying the quadratic formula involves a few straightforward steps:
  • Identify the coefficients \( a \), \( b \), and \( c \).
  • Calculate the discriminant \( b^2 - 4ac \).
  • Use the quadratic formula to find the roots.
Remember, the "\( \pm \)" symbol in the formula indicates that you will typically obtain two solutions or "roots" for the quadratic equation, unless the value under the square root is zero.
Discriminant
Understanding the discriminant is crucial when solving quadratic equations, as it tells us the nature of the roots. The discriminant is the part of the quadratic formula located under the square root: \( b^2 - 4ac \). It plays a significant role in determining whether the solutions of a quadratic equation are real, repeated, or imaginary.

Here’s how different values of the discriminant affect the solutions:
  • If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
  • If \( b^2 - 4ac = 0 \), there is exactly one real solution (a repeated root).
  • If \( b^2 - 4ac < 0 \), there are no real solutions; instead, the solutions are complex or imaginary numbers.
In the step-by-step solution for this exercise, you calculated the discriminant for each equation. For example:
  • In the first equation, the discriminant was negative (\( 144 - 1920 \)), leading to no real solutions.
  • However, the second equation had a positive discriminant (\( 900 - 240 \)), indicating two real solutions are possible.
Real Solutions
Real solutions refer to solutions that are not complex or imaginary numbers. When solving quadratic equations, determining whether the solutions are real depends on the discriminant. As mentioned in the previous section, when the discriminant is positive, it indicates that the equation has two distinct real solutions.

Real solutions are important in many real-world applications as they often represent measurable quantities or dimensions. For instance, in the step-by-step example, one of the quadratic equations had real solutions, thanks to a positive discriminant. This meant you could find actual values of \( t \) that satisfy the equation.

To ensure successful calculation of real solutions using the quadratic formula:
  • Ensure the discriminant is correctly computed.
  • Only proceed with finding the square root if the discriminant is non-negative.
  • Substitute the values into the quadratic formula to get the accurate real solutions.
Understanding when and how real solutions occur helps clarify the results of a quadratic equation and confirms their applicability to real-life scenarios.

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Most popular questions from this chapter

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