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Chapter 7: Problem 26

Solve. If no solution exists, state this. $$t+\frac{6}{t}=-5$$

Short Answer

Expert verified
The solutions are \( t = -2 \) and \( t = -3 \).

Step by step solution

01

Multiply through by t

To eliminate the fraction, multiply every term by t: \[ t \times t + t \times \frac{6}{t} = -5t \] which simplifies to: \[ t^2 + 6 = -5t \]
02

Form a quadratic equation

Rearrange the equation to standard quadratic form (ax^2 + bx + c = 0): \[ t^2 + 5t + 6 = 0 \]
03

Factor the quadratic equation

Factor the quadratic equation: \[ (t + 2)(t + 3) = 0 \]
04

Solve for t

Find the values of t that satisfy the equation. Set each factor equal to zero: \[ t + 2 = 0 \] which gives: \[ t = -2 \] and \[ t + 3 = 0 \] which gives: \[ t = -3 \]
05

Verify the solutions

Substitute each solution back into the original equation to verify:For \( t = -2 \):\[ -2 + \frac{6}{-2} = -2 - 3 = -5 \] which is true.For \( t = -3 \):\[ -3 + \frac{6}{-3} = -3 - 2 = -5 \] which is also true.

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Key Concepts

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

Multiplying Equations
Multiplying equations can help simplify problems involving fractions. When you see a fraction in an equation, you can eliminate it by multiplying every term by the same value.
For example, in the given exercise, we start with:
\[ t + \frac{6}{t} = -5 \]
To make the fraction disappear, multiply each term by \( t \):
\[ t \times t + t \times \frac{6}{t} = -5t \]
This gives us:
\[ t^2 + 6 = -5t \]
Now, we no longer have a fraction.
Factoring Quadratics
Factoring is a method used to solve quadratic equations. A quadratic equation can be in the form \( ax^2 + bx + c = 0 \). Here, \(a\), \(b\), and \(c\) are constants.
For our example, after multiplying we had:
\[ t^2 + 6 = -5t \]
We rearrange to get standard quadratic form:
\[ t^2 + 5t + 6 = 0 \]
Next, we factor the left-hand side:
\[ (t + 2)(t + 3) = 0 \]
This means if either \( t + 2 = 0 \) or \( t + 3 = 0 \), the equation holds.
Solving for \( t \), we get two possible solutions: \( t = -2 \) and \( t = -3 \).
Verifying Solutions
Verifying solutions ensures our answers make sense in the original equation.
To do this, substitute the solutions back into the initial equation.
First, check \( t = -2 \):
\[ -2 + \frac{6}{-2} = -2 - 3 = -5 \]
This verifies that \( t = -2 \) is indeed a solution.
Now, check \( t = -3 \):
\[ -3 + \frac{6}{-3} = -3 - 2 = -5 \]
This confirms that \( t = -3 \) is also a solution.
Verifying solutions is crucial because it verifies the correctness of our answers.
Standard Quadratic Form
Standard quadratic form is an important concept you need to know. A quadratic equation generally looks like:
\[ ax^2 + bx + c = 0 \]
Here, \(a\), \(b\), and \(c\) are constants and x is the variable.
In our example, we started with:
\[ t^2 + 6 = -5t \]
We rearranged it to fit the standard quadratic form:
\[ t^2 + 5t + 6 = 0 \]
The constants are \(a = 1\), \(b = 5\), and \(c = 6\).
Knowing the standard form helps you recognize patterns and use factoring methods to solve the equation efficiently.

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