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Chapter 8: Problem 11

Solve each equation. Check the solutions. $$3-\frac{1}{t}=\frac{2}{t^{2}}$$

Short Answer

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

Step by step solution

01

Identify the common denominator

The common denominator for the terms involving t is \(t^2\).
02

Multiply both sides by the common denominator

Multiply both sides of the equation by \(t^2\) to eliminate the fractions: \[ t^2 \times \bigg(3 - \frac{1}{t} \bigg) = t^2 \times \frac{2}{t^2} \] which simplifies to \[ 3t^2 - t = 2 \]
03

Move all terms to one side of the equation

Rearrange the equation to form a standard quadratic equation: \[ 3t^2 - t - 2 = 0 \]
04

Solve the quadratic equation

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \(a = 3\), \(b = -1\), and \(c = -2\): \[ t = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 3 \times (-2)}}{2 \times 3} \] which simplifies to \[ t = \frac{1 \pm \sqrt{1 + 24}}{6} \] and further simplifies to \[ t = \frac{1 \pm 5}{6} \]. This gives two potential solutions: \[ t = 1 \] and \[ t = -\frac{2}{3} \].
05

Check the solutions

Substitute \(t = 1\) back into the original equation: \[ 3 - \frac{1}{1} = \frac{2}{1^2} \] which simplifies to \[ 2 = 2 \]. Therefore, \(t = 1\) is a valid solution.Substitute \(t = -\frac{2}{3}\) back into the original equation: \[ 3 - \frac{1}{-\frac{2}{3}} = \frac{2}{\big( -\frac{2}{3} \big)^2} \] which simplifies to \[ 3 + \frac{3}{2} = \frac{2}{\frac{4}{9}} \] and further simplifies to \[ \frac{9}{2} = \frac{9}{2} \]. Therefore, \(t = -\frac{2}{3}\) is also a valid solution.

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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 to solve quadratic equations of the form \(ax^2 + bx + c = 0\). A quadratic equation involves variables raised to the power of 2. The standard method for solving them involves using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here鈥檚 a breakdown of each part:
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Most popular questions from this chapter

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