Problem 65 - Algebra for College Students

Question

Find all real solutions to each equation. $$x^{-2}-2 x^{-1}-1=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \sqrt{2} - 1 \) and \( x = -1 - \sqrt{2} \).

Step 1: Rewrite the equation

Start by letting a new variable represent the term with the negative exponent. Set Let \(y = x^{-1}\), which simplifies to \( y = \frac{1}{x} \). Substitute \( y\) into the equation: \(y^2 - 2y - 1 = 0\).

Step 2: Solve the quadratic equation

The equation can now be solved by the quadratic formula where \(a = 1\), \(b = -2\), and \(c = -1\). Use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Substitute the values:\[ y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} \] \[ y = \frac{2 \pm \sqrt{4 + 4}}{2} \] \[ y = \frac{2 \pm \sqrt{8}}{2} \] \[ y = \frac{2 \pm 2\sqrt{2}}{2} \] \[ y = 1 \pm \sqrt{2} \].

Step 3: Return to original variable

Recall \( y = \frac{1}{x} \). Therefore, \( x = \frac{1}{y} \). We have two solutions from Step 2: \( y = 1 + \sqrt{2} \Rightarrow x = \frac{1}{1 + \sqrt{2}} \) \( y = 1 - \sqrt{2} \Rightarrow x = \frac{1}{1 - \sqrt{2}} \).

Step 4: Simplify the solutions

Rationalize the denominators for the solutions: For \(x = \frac{1}{1 + \sqrt{2}}\): \[ x = \frac{1}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}}{1 - \sqrt{2}} = \frac{1 - \sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})} = \frac{1 - \sqrt{2}}{1 - 2} = \frac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1 \]. For \(x = \frac{1}{1 - \sqrt{2}}\): \[ x = \frac{1}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + \sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})} = \frac{1 + \sqrt{2}}{1 - 2} = \frac{1 + \sqrt{2}}{-1} = -1 - \sqrt{2} \].

Key Concepts

Negative Exponents

When dealing with negative exponents, it's essential to understand that they represent the reciprocal of the base with a positive exponent. For example, the expression \( x^{-1} \) is equivalent to \( \frac{1}{x} \). This is a key step in transforming complex equations into more manageable forms. In the given exercise, you convert \( x^{-2} \) to \( y^2 \) by letting \( y = x^{-1} \). This substitution transforms the original equation into a simpler quadratic form.

Substitution Method

The substitution method is a powerful technique used in solving equations, particularly when they involve complex expressions or multiple variables. Here, by setting \( y = x^{-1} \), we replace \( x^{-2} \) with \( y^2 \) and \( x^{-1} \) with \( y \). This simplification helps us convert the difficult equation \( x^{-2} - 2x^{-1} - 1 = 0 \) into a more familiar form: \( y^2 - 2y - 1 = 0 \). Once simplified, it becomes much easier to solve using standard methods.

Quadratic Formula

The quadratic formula is a foolproof method for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). The formula is \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. To apply this formula, identify the coefficients \( a, b, \) and \( c \). In our equation \( y^2 - 2y - 1 = 0 \), we have \( a = 1 \), \( b = -2 \), and \( c = -1 \). Substituting these values into the formula yields: \[ y = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2} \]. This provides two solutions for \( y \), which are then re-substituted back to find the solutions for \( x \).

Rationalizing Denominators

Rationalizing the denominator means rewriting a fraction so the denominator contains no radicals. This step often simplifies further calculations and makes the result more readable. For the solution \( x = \frac{1}{1 + \sqrt{2}} \), multiplying the numerator and the denominator by the conjugate \( 1 - \sqrt{2} \) helps eliminate the radical in the denominator: \[ x = \frac{1(1 - \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{1 - \sqrt{2}}{1 - 2} = \sqrt{2} - 1 \]. Similarly, for \( x = \frac{1}{1 - \sqrt{2}} \), rationalizing by multiplying by \( 1 + \sqrt{2} \) yields: \[ x = \frac{1(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})} = \frac{1 + \sqrt{2}}{1 - 2} = -1 - \sqrt{2} \]. This process ensures the solutions are in their simplest and most understandable form.