Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 4+\frac{1}{x}-\frac{1}{x^{2}}=0 $$

Short answer

The real solutions are \( x = \frac{-1 + \sqrt{17}}{8} \) and \( x = \frac{-1 - \sqrt{17}}{8} \).

Step-by-step solution

Rewrite the Equation

Rewrite the given equation in a standard quadratic form. Multiply every term by \( x^2 \) to eliminate the denominators.

Simplify the Equation

After multiplying by \( x^2 \): \( x^2 (4) + x^2 \frac{1}{x} - x^2 \frac{1}{x^2} = 0 \), which simplifies to: \( 4x^2 + x - 1 = 0 \)

Identify Coefficients for the Quadratic Formula

Identify the coefficients \( a \), \( b \), and \( c \) in the quadratic equation: \( a = 4 \), \( b = 1 \), and \( c = -1 \)

Apply the Quadratic Formula

Recall the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the coefficients into the formula.

Calculate the Discriminant

\( b^2 - 4ac = (1)^2 - 4(4)(-1) = 1 + 16 = 17 \). The discriminant is 17.

Solve for the Roots

Using the quadratic formula: \( x = \frac{-1 \pm \sqrt{17}}{8} \). Therefore, the solutions are: \( x = \frac{-1 + \sqrt{17}}{8} \) and \( x = \frac{-1 - \sqrt{17}}{8} \).

Key concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree 2. The general form of a quadratic equation is:
\[ ax^2 + bx + c = 0 \]
where:

• \( a \) is the coefficient of \( x^2 \), and it cannot be zero.
• \( b \) is the coefficient of \( x \).
• \( c \) is the constant term.

In the given problem, the original equation is:
\[ 4 + \frac{1}{x} - \frac{1}{x^2} = 0 \]
First, to simplify this into a standard quadratic equation, you multiply every term by \( x^2 \) to eliminate the denominators. This will give us: \[ 4x^2 + x - 1 = 0 \]
Now you can see the equation has taken the form \( ax^2 + bx + c = 0 \). Here, \( a = 4 \), \( b = 1 \), and \( c = -1 \).

Discriminant

The discriminant is a crucial part of the quadratic formula. It helps determine the nature of the solutions of a quadratic equation.
The discriminant is given by the expression:
\[ \text{Discriminant} = b^2 - 4ac \]
Depending on its value, it indicates:

• If it’s positive (> 0): There are two distinct real solutions.
• If it’s zero (0): There is exactly one real solution, also known as a repeated or double root.
• If it’s negative (< 0): There are no real solutions, but two complex solutions.

In our case, let’s substitute the values of \( a \), \( b \), and \( c \): \[ b^2 - 4ac = (1)^2 - 4(4)(-1) = 1 + 16 = 17 \]
Since the discriminant here is 17, which is greater than 0, the equation has two distinct real solutions.

Solving Equations

The quadratic formula is a universal method for solving quadratic equations. It’s expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Let's use this formula to solve our simplified equation \( 4x^2 + x - 1 = 0 \). We already have the coefficients:

• \( a = 4 \)
• \( b = 1 \)
• \( c = -1 \)

Substitute them into the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{17}}{8} \]
This simplifies to two solutions:
\[ x = \frac{-1 + \sqrt{17}}{8} \]
\[ x = \frac{-1 - \sqrt{17}}{8} \]
This tells us the quadratic equation has two real and distinct solutions, which were obtained by plugging in the values of \( a \), \( b \), and \( c \) back into the quadratic formula.
Remember, the \( \pm \) sign in the formula is what gives us the two different solutions.