Problem 68 - Algebra for College Students

Question

Solve each equation by an appropriate method. $$\frac{1}{x^{2}}+\frac{3}{x}+1=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-3 + \sqrt{5}}{2}\) and \(x = \frac{-3 - \sqrt{5}}{2}\).

Step 1: Identify Type of Equation

Recognize that the equation is a quadratic equation in terms of the variable \(\frac{1}{x}\). Rewrite the variable \(\frac{1}{x}\) as \(y\), so the equation becomes \(\frac{1}{y^{2}} + \frac{3}{y} + 1 = 0\).

Step 2: Multiply by \(y^{2}\)

To simplify the equation, multiply every term by \(y^{2}\) to eliminate the fraction: \((y^{2} * \frac{1}{y^{2}}) + (y^{2} * \frac{3}{y}) + y^{2} * 1 = 0\) which simplifies to \(1 + 3y + y^{2} = 0\).

Step 3: Write in Standard Quadratic Form

Rewrite the equation in the standard quadratic form: \(y^{2} + 3y + 1 = 0\).

Step 4: Use the Quadratic Formula

Use the quadratic formula \(y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\) where \(a = 1\), \(b = 3\), and \(c = 1\). Substitute these values into the formula: \(y = \frac{{-3 \pm \sqrt{{3^2 - 4*1*1}}}}{{2*1}}\).

Step 5: Simplify the Roots

Simplify under the square root and solve: \(y = \frac{{-3 \pm \sqrt{9 - 4}}}{{2}}\) which simplifies to \(y = \frac{{-3 \pm \sqrt{5}}}{{2}}\).

Step 6: Substitute Back \(y = \frac{1}{x}\)

Substitute \(y = \frac{1}{x}\) back into the solved values: \(\frac{1}{x} = \frac{{-3 + \sqrt{5}}}{{2}}\) and \(\frac{1}{x} = \frac{{-3 - \sqrt{5}}}{{2}}\).

Step 7: Solve for \(x\)

Take the reciprocals to solve for \(x\): \(x = \frac{2}{{-3 + \sqrt{5}}}\) and \(x = \frac{2}{{-3 - \sqrt{5}}}\). Rationalize the denominators (multiply the numerator and denominator by the conjugate of the denominator): \(x = \frac{2(-3 - \sqrt{5})}{(-3 + \sqrt{5})(-3 - \sqrt{5})}\) and \(x = \frac{2(-3 + \sqrt{5})}{(-3 - \sqrt{5})(-3 + \sqrt{5})}\).

Step 8: Further Simplify

Simplify the rationalization: \(x = \frac{-6 - 2 \sqrt{5}}{9 - 5} \Rightarrow x = \frac{-6 - 2 \sqrt{5}}{4} \Rightarrow x = \frac{-3 - \sqrt{5}}{2}\) and \(x = \frac{-6 + 2 \sqrt{5}}{4} \Rightarrow x = \frac{-3 + \sqrt{5}}{2}\).

Key Concepts

quadratic formula

The quadratic formula is a powerful tool used to solve quadratic equations. A quadratic equation is generally in the form of \( ax^2 + bx + c = 0 \). The quadratic formula is given by \( x = \frac{{-b \pm \sqrt{b^2 - 4ac}}}{2a} \). This formula helps you find the values of \(x\) that satisfy the equation. It is derived from the process of completing the square and works for any quadratic equation. To use it, you need to identify the coefficients \(a\), \(b\), and \(c\) from your equation.

variable substitution

Variable substitution can simplify complex equations. In this exercise, we start with the equation \(\frac{1}{x^2} + \frac{3}{x} + 1 = 0\). Since \(\frac{1}{x}\) appears more than once, replacing \(\frac{1}{x}\) with a new variable like \(y\) can make the equation easier to handle. We let \( y = \frac{1}{x} \), transforming the equation into \(y^2 + 3y + 1 = 0\). This new form is a more recognizable quadratic equation, allowing us to use familiar techniques to solve it.

solving equations

Solving equations involves finding the values of variables that make the equation true. For the quadratic equation \( y^2 + 3y + 1 = 0\), we use the quadratic formula. First, identify \(a\), \(b\), and \(c\) where \(a = 1\), \(b = 3\), and \(c = 1\). Substitute these values into the formula: \( y = \frac{{-3 \pm \sqrt{9 - 4}}}{2} \). This simplifies to \( y = \frac{{-3 \pm \sqrt{5}}}{2} \). After solving for \(y\), substitute back \( y = \frac{1}{x} \) to find \( x \). Finally, solve the resulting expressions \( \frac{1}{x} = \frac{{-3 + \sqrt{5}}}{2} \) and \( \frac{1}{x} = \frac{{-3 - \sqrt{5}}}{2}\).

rationalizing denominators

Rationalizing denominators is a method used to eliminate radicals from the denominator of a fraction. In the final step of this exercise, we have fractions like \( x = \frac{2}{{-3 + \sqrt{5}}} \). To rationalize, we multiply both the numerator and the denominator by the conjugate of the denominator: \( -3 - \sqrt{5} \). This turns \( x = \frac{2(-3 - \sqrt{5})}{(-3 + \sqrt{5})(-3 - \sqrt{5})} \). Simplifying this gives \( x = \frac{-6 - 2 \sqrt{5}}{4} \). This process helps to write the solution in a more standard form that is easier to work with and understand.