/*! 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 31 Solve. $$4 t^{-2}-3 t^{-1}-1=0... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 31

Solve. $$4 t^{-2}-3 t^{-1}-1=0$$

Short Answer

Expert verified
The solutions are \( t = 1 \) and \( t = -4 \).

Step by step solution

01

Substitute variables to simplify

Let us set a new variable to simplify the equation. Let鈥檚 use \( u = t^{-1} \), thus \( t^{-2} = u^2 \). Therefore, the equation becomes \( 4u^2 - 3u - 1 = 0 \).
02

Solve the quadratic equation

Now, solve the quadratic equation \( 4u^2 - 3u - 1 = 0 \) using the quadratic formula, \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 4 \), \( b = -3 \), and \( c = -1 \).
03

Apply the quadratic formula

Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula: \[ u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-1)}}{2 \cdot 4} \]. Simplifying inside the square root, this becomes \[ u = \frac{3 \pm \sqrt{9 + 16}}{8} \]. Hence, \[ u = \frac{3 \pm 5}{8} \].
04

Determine the two solutions for u

Calculate the two values for \( u \): \[ u_1 = \frac{3 + 5}{8} = 1 \] and \[ u_2 = \frac{3 - 5}{8} = -\frac{1}{4} \]. Thus, \( u = 1 \) or \( u = -\frac{1}{4} \).
05

Back-substitute the original variable

Recall that \( u = t^{-1} = \frac{1}{t} \). Therefore, \( t = 1 \) when \( u = 1 \), and \( t = -4 \) when \( u = -\frac{1}{4} \).
06

Verify the solutions

Substitute \( t = 1 \) and \( t = -4 \) back into the original equation \( 4t^{-2} - 3t^{-1} - 1 = 0 \) to verify the solutions. Both should satisfy the equation.

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Key Concepts

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

Quadratic Equations
A quadratic equation is a second-degree polynomial equation in a single variable. The general form is: \[ ax^2 + bx + c = 0 \] where:
  • \( a \), \( b \), and \( c \) are constants, where \( a eq 0 \)
  • \( x \) is the variable
Quadratic equations can have two solutions (real or complex). These solutions are the values of \( x \) that satisfy the equation. To solve these equations, various methods like factoring, completing the square, and using the quadratic formula can be employed.
Variable Substitution
Variable substitution is a powerful technique used to simplify complex equations. It involves replacing complex expressions with a simpler variable.
In the exercise, we used the substitution: \[ u = t^{-1} \]
This transformed the original equation \( 4 t^{-2} - 3 t^{-1} - 1 = 0 \) into a simpler quadratic equation:
\( 4u^2 - 3u - 1 = 0 \) This makes it easier to solve, as we recognize it as a standard quadratic equation.
Quadratic Formula
The quadratic formula is a universal method for solving any quadratic equation. The formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where:
  • \( a \), \( b \), and \( c \) are coefficients from the equation \( ax^2 + bx + c = 0 \)
  • The symbol \( \pm \) indicates there are generally 2 solutions, one for addition and one for subtraction
Let's summarize the usage of this formula through the given exercise. For the equation \( 4u^2 - 3u - 1 = 0 \), the coefficients are:
  • \( a = 4 \)
  • \( b = -3 \)
  • \( c = -1 \)
Substituting these into the quadratic formula yields: \[ u = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 4 \cdot (-1)}}{2\cdot 4} \] Simplifying this step by step, we get: \[ u = \frac{3 \pm \sqrt{9 + 16}}{8} \] \[ u = \frac{3 \pm 5}{8} \] Therefore the solutions are:
  • \( u_1 = \frac{3 + 5}{8} = 1 \)
  • \( u_2 = \frac{3 - 5}{8} = -\frac{1}{4} \)
These values can then be back-substituted as needed to revert to the original variable of the problem.

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