Problem 66 - Algebra for College Students

Question

Find all real solutions to each equation. $$x^{-2}-6 x^{-1}+6=0$$

Step-by-Step Solution

Answer

The real solutions are \( x = \frac{1}{2} - \frac{\sqrt{3}}{6} \) and \( x = \frac{1}{2} + \frac{\sqrt{3}}{6} \).

Step 1 - Identify the substitution

To simplify the given equation, substitute a new variable. Let us define a new variable: \[ y = x^{-1} \].

Step 2 - Rewrite the equation

Substitute the new variable into the original equation:\[ y^2 - 6y + 6 = 0 \]. This transforms the equation into a standard quadratic form.

Step 3 - Solve the quadratic equation

Use the quadratic formula to solve the quadratic equation. The quadratic formula is:\[ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]. For our equation, \(a = 1\), \(b = -6\), and \(c = 6\).Substitute these values into the quadratic formula:\[ y = \frac{{6 \pm \sqrt{{36 - 24}}}}{2} = \frac{{6 \pm \sqrt{12}}}{2} = \frac{{6 \pm 2\sqrt{3}}}{2} = 3 \pm \sqrt{3} \]

Step 4 - Back-substitute into the original variable

Recall that \(y = x^{-1}\), which implies \(\frac{1}{x} = 3 + \sqrt{3}\) and \(\frac{1}{x} = 3 - \sqrt{3}\). Solve for \(x\):\[ x = \frac{1}{3 + \sqrt{3}} \] and \[ x = \frac{1}{3 - \sqrt{3}} \]

Step 5 - Rationalize the denominators

To simplify the expressions for \(x\), rationalize the denominators:\[ x = \frac{1}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{3 - \sqrt{3}}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3 - \sqrt{3}}{9 - 3} = \frac{3 - \sqrt{3}}{6} = \frac{3}{6} - \frac{\sqrt{3}}{6} = \frac{1}{2} - \frac{\sqrt{3}}{6} \]Similarly,\[ x = \frac{1}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 + \sqrt{3}}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{3 + \sqrt{3}}{9 - 3} = \frac{3 + \sqrt{3}}{6} = \frac{3}{6} + \frac{\sqrt{3}}{6} = \frac{1}{2} + \frac{\sqrt{3}}{6} \]

Key Concepts

Algebraic Substitution

Algebraic substitution is a method used to simplify complex equations by introducing a new variable. For instance, in our given problem, we start with the equation:

\[ x^{-2} - 6x^{-1} + 6 = 0 \]Using algebraic substitution, we let a new variable replace a part of the equation that repeats or makes the equation easier to handle. Here, we set:\[ y = x^{-1} \]This transformation changes our equation into a more manageable quadratic form:\[ y^2 - 6y + 6 = 0 \]Now the equation looks simpler and can be solved using standard quadratic techniques. This method is very helpful because it turns difficult problems into ones we already know how to solve. Once solved, we substitute back into our original variable to get the final solution.

Quadratic Formula

The quadratic formula is a powerful tool to find the solutions of quadratic equations. The general form of a quadratic equation is:\[ ax^2 + bx + c = 0 \]The quadratic formula to solve for \(x\) is:\[ x = \frac{{-b \,\pm\, \sqrt{{b^2 - 4ac}}}}{2a} \]For our substituted variable equation, \( y^2 - 6y + 6 = 0 \), we can see that:\[ a = 1, \quad b = -6, \quad c = 6 \]Substituting these values into the quadratic formula gives us:\[ y = \frac{{-(-6) \,\pm\, \sqrt{{(-6)^2 - 4(1)(6)}}}}{2(1)} = \frac{{6 \,\pm\, \sqrt{{36 - 24}}}}{2} = \frac{{6 \,\pm\, \sqrt{12}}}{2} \]Further simplifying the root term, we get:\[ y = \frac{{6 \,\pm\, 2\sqrt{3}}}{2} \]This is reduced to:\[ y = 3 \,\pm\, \sqrt{3} \]Thus, the solutions for \(y\) are \( y = 3 + \sqrt{3} \) and \( y = 3 - \sqrt{3} \). Using these results, we can now back-substitute to find the original variable \(x\).

Rationalizing Denominators

Rationalizing the denominator involves removing any irrational numbers from the denominator of a fraction. After back-substituting, we get:\[ \frac{1}{x} = 3 + \sqrt{3} \quad \text{and} \quad \frac{1}{x} = 3 - \sqrt{3} \]To solve for \( x \), we take the reciprocal but first need to rationalize the denominators:For \( \frac{1}{3 + \sqrt{3}} \), multiply by the conjugate \( \frac{3 - \sqrt{3}}{3 - \sqrt{3}} \):\[ x = \frac{1}{3 + \sqrt{3}} \cdot \frac{3 - \sqrt{3}}{3 - \sqrt{3}} = \frac{3 - \sqrt{3}}{(3 + \sqrt{3})(3 - \sqrt{3})} = \frac{3 - \sqrt{3}}{9 - 3} = \frac{3 - \sqrt{3}}{6} = \frac{1}{2} - \frac{\sqrt{3}}{6} \]Similarly, for \( \frac{1}{3 - \sqrt{3}} \), multiply by the conjugate \( \frac{3 + \sqrt{3}}{3 + \sqrt{3}} \):\[ x = \frac{1}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 + \sqrt{3}}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{3 + \sqrt{3}}{9 - 3} = \frac{3 + \sqrt{3}}{6} = \frac{1}{2} + \frac{\sqrt{3}}{6} \]This step ensures our final answers are in a much simpler form, removing radicals from the denominator for better clarity and usability.