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Chapter 9: Problem 191

Solve. $$8 x^{-2}-2 x^{-1}-3=0$$

Short Answer

Expert verified
The solutions are \( x = \frac{4}{3} \) or \( x = -2 \).

Step by step solution

01

Substitute variables

Let us simplify the expression by substituting the variables. Let:\[ y = x^{-1} \]Thus, the equation \( 8x^{-2} - 2x^{-1} - 3 = 0 \) becomes:\[ 8y^2 - 2y - 3 = 0 \]
02

Solve the quadratic equation

Now solve the quadratic equation formed:\[ 8y^2 - 2y - 3 = 0 \]The quadratic formula is:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a = 8 \), \( b = -2 \), and \( c = -3 \). Substituting these values, we get:\[ y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 8 \cdot (-3)}}{2 \cdot 8} = \frac{2 \pm \sqrt{4 + 96}}{16} = \frac{2 \pm \sqrt{100}}{16} \]Further simplifying:\[ y = \frac{2 \pm 10}{16} \]Thus, the solutions for \( y \) are:\[ y = \frac{12}{16} = \frac{3}{4} \quad \text{or} \quad y = \frac{-8}{16} = -\frac{1}{2} \]
03

Back-substitute the original variable

Now revert back to the original variable \( x \). Recall that \( y = x^{-1} \), or \( y = \frac{1}{x} \). Therefore, we have:\[ \frac{1}{x} = \frac{3}{4} \quad \text{or} \quad \frac{1}{x} = -\frac{1}{2} \]This gives us:\[ x = \frac{4}{3} \quad \text{or} \quad x = -2 \]

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Key Concepts

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

Variable Substitution
In mathematics, variable substitution is a method used to simplify equations. The idea is to replace a complex part of the equation with a new variable. This makes the equation easier to solve.
For instance, consider the equation: \[ 8x^{-2} - 2x^{-1} - 3 = 0 \]
Here, we substitute \[ y = x^{-1} \]
Now, the equation becomes simpler:
\[ 8y^2 - 2y - 3 = 0 \]
By using a simpler variable, we transform a complex problem into an easier one to handle. Once we solve the simpler equation, we can revert back to the original variable.
Solving Quadratic Equations
A quadratic equation is an equation of the form \[ ax^2 + bx + c = 0 \]
In this problem, after substitution, our equation is:
\[ 8y^2 - 2y - 3 = 0 \]
To solve this, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Identify the coefficients as follows:
    \[ a = 8 \] \[ b = -2 \] \[ c = -3 \]
Substitute these into the quadratic formula:
\[ y = \frac{2 \pm \sqrt{4 + 96}}{16} \]
This simplifies to:
\[ y = \frac{2 \pm 10}{16} \]
So the solutions for y are:
    \[ y = \frac{12}{16} = \frac{3}{4} \] \[ y = \frac{-8}{16} = -\frac{1}{2} \]
These steps provide a clear path to finding solutions in a quadratic equation.
Rational Exponents
Understanding rational exponents can simplify many math problems. A rational exponent is an exponent that is a fraction. For instance, \[ x^{-1} \] is the same as saying \[ \frac{1}{x} \]
In the original problem, we have terms like \[ x^{-2} \] and \[ x^{-1} \]
These can be transformed using rational exponents:
  • \[ x^{-2} = \frac{1}{x^2} \]
  • \[ x^{-1} = \frac{1}{x} \]
By treating these as \[ y^2 \] and \[ y \], the equation becomes easier to solve. After solving for y, don't forget to revert to the original variable x. For example:
\[ y = \frac{3}{4} \] translates to \[ \frac{1}{x} = \frac{3}{4} \]
Which means \[ x = \frac{4}{3} \]
This way, rational exponents help bridge complex problems into more understandable parts.

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Most popular questions from this chapter

Solve each inequality algebraically and write any solution in interval notation. $$-2 x^{2}+8 x-10<0$$

Find the intercepts of the parabola whose function is given. $$f(x)=-3 x^{2}+x-1$$

(a) rewrite each function in \(f(x)=a(x-h)^{2}+k\) form and (b) graph it by using transformations. $$f(x)=-2 x^{2}+8 x-10$$

(a) solve graphically and (b) write the solution in interval notation. $$x^{2}+6 x+5>0$$

Find the intercepts of the parabola whose function is given. $$f(x)=x^{2}+8 x+12$$
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