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Chapter 8: Problem 26

Solve. (Find all complex-number solutions.) \(15 t^{2}+7 t=2\)

Short Answer

Expert verified
The solutions are \( t = \frac{1}{5} \) and \( t = -\frac{2}{3} \).

Step by step solution

01

Rewrite the Equation in Standard Form

First, rewrite the given quadratic equation in the standard form, which is \( at^2 + bt + c = 0 \).Starting with:\( 15t^2 + 7t = 2 \)Subtract 2 from both sides to get:\( 15t^2 + 7t - 2 = 0 \)
02

Identify Coefficients

Identify the coefficients from the standard form equation, which are:\( a = 15 \), \( b = 7 \), \( c = -2 \)
03

Apply the Quadratic Formula

Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)Substitute the identified coefficients into the formula:\( t = \frac{-7 \pm \sqrt{7^2 - 4(15)(-2)}}{2(15)} \)This simplifies to:\( t = \frac{-7 \pm \sqrt{49 + 120}}{30} \)Simplify further:\( t = \frac{-7 \pm \sqrt{169}}{30} \)
04

Solve for Roots

Simplify the square root and solve for the two possible values of t:\( t = \frac{-7 + 13}{30} \), and \( t = \frac{-7 - 13}{30} \)So the roots are:\( t = \frac{6}{30} = \frac{1}{5} \), and \( t = \frac{-20}{30} = -\frac{2}{3} \)

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Key Concepts

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

Standard Form Equation
To solve a quadratic equation, it must first be in the standard form. The standard form of a quadratic equation is expressed as: \( at^2 + bt + c = 0 \)Here,
  • \(a\) represents the coefficient of the term with the variable squared (\(t^2\))
  • \(b\) is the coefficient of the term with the variable to the first power (\(t\))
  • \(c\) is the constant term
For example, the equation \( 15t^2 + 7t = 2 \), needs to be rewritten in the standard form. Subtract 2 from both sides to get: \( 15t^2 + 7t - 2 = 0 \)This transformation allows you to identify the coefficients more easily.
Quadratic Formula
The quadratic formula is a powerful tool for solving equations that are in the standard form. The formula is given by: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where:
  • \(a\), \(b\), and \(c\) are the coefficients from the standard form equation
  • \(\pm\) indicates that there are generally two possible solutions, known as the roots
This formula will find the solutions to any quadratic equation as long as you correctly substitute the values of \(a\), \(b\), and \(c\). In the given problem, after transforming it to standard form, we identified:
  • \(a = 15\)
  • \(b = 7\)
  • \(c = -2\)
Substituting these into the quadratic formula, we get: \[ t = \frac{-7 \pm \sqrt{49 + 120}}{30} \]Solve further to simplify and find the roots.
Complex-Number Solutions
In quadratic equations, sometimes the discriminant (the part under the square root sign, \(b^2 - 4ac\)) may be negative. If that's the case, the equation has complex-number solutions. Complex numbers include the imaginary unit \(i\), where \( i^2 = -1 \).If the discriminant is positive or zero, the solutions will be real numbers, as in this exercise:
  • The discriminant is \(49 + 120 = 169\), which is positive
  • Thus, the solutions are real numbers
When the discriminant is negative, you would use the imaginary unit \(i\) to express the solutions:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \rightarrow t = \frac{-b \pm \sqrt{-x}}{2a} \rightarrow t = \frac{-b \pm i\sqrt{x}}{2a} \]Here, \(x\) is a positive number, and the \(i\) indicates that the solutions are complex.
Solving Quadratic Equations
Solving quadratic equations requires careful handling of algebraic steps. Let's piece together the problem:Starting from our standard form equation, \(15t^2 + 7t - 2 = 0\):
  1. Use the quadratic formula, substituting the coefficients:
    2. \( t = \frac{-7 \pm \sqrt{169}}{30} \)
  2. Simplify the square root:
    3. \(\sqrt{169} = 13 \)
  3. Split into two cases for the \(\pm\) sign:
    • \( t = \frac{-7 + 13}{30} = \frac{6}{30} = \frac{1}{5} \)
    • \( t = \frac{-7 - 13}{30} = \frac{-20}{30} = -\frac{2}{3} \)
Therefore, the solutions to \(15 t^2 + 7 t - 2 = 0\) are \( t = \frac{1}{5} \) and \( t = -\frac{2}{3} \). Breaking down these steps carefully and substituting accurately makes solving quadratic equations straightforward.

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