Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ \frac{3 x}{x-2}+\frac{1}{x}=4 $$

Short answer

The real solutions are: \( x = \frac{9 + \sqrt{73}}{2} \) and \( x = \frac{9 - \sqrt{73}}{2} \)

Step-by-step solution

Clear the fractions

To clear the fractions, find the common denominator, which is \(x(x-2)\). Multiply every term by \(x(x-2)\): \[ \frac{3x}{x-2} \times x(x-2) + \frac{1}{x} \times x(x-2) = 4 \times x(x-2) \]

Simplify the equation

Simplify each term: \[ 3x^2 + (x-2) = 4x^2 - 8x \]

Combine like terms

Combine all the terms to one side of the equation: \[ 3x^2 + x - 2 = 4x^2 - 8x \]Move all terms to one side to set the equation to 0: \[ 0 = 4x^2 - 8x - 3x^2 - x + 2 \]Simplify: \[ 0 = x^2 - 9x + 2 \]

Apply the quadratic formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -9 \), and \( c = 2 \): \[ x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(2)}}{2(1)} \]

Simplify within the quadratic formula

Simplify the terms under the square root and the rest of the formula: \[ x = \frac{9 \pm \sqrt{81 - 8}}{2} = \frac{9 \pm \sqrt{73}}{2} \]

Write the final solutions

The real solutions are: \[ x = \frac{9 + \sqrt{73}}{2} \text{ and } x = \frac{9 - \sqrt{73}}{2} \]

Key concepts

Solve quadratic equations

To solve quadratic equations, one powerful and universal method we use is the quadratic formula. A quadratic equation typically has the form: \[ ax^2 + bx + c = 0 \] The quadratic formula helps us find the values of \( x \) that satisfy this equation. The formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a \), \( b \), and \( c \) are coefficients from the equation.
To use the quadratic formula, follow these steps:

• Identify \( a \), \( b \), and \( c \) from the quadratic equation.
• Substitute these values into the formula.
• Simplify the expression under the square root (known as the discriminant).
• Calculate the values of \( x \) using the formula.

This method works for all quadratic equations, whether they have real or complex solutions.

Real solutions

When dealing with quadratic equations, the nature of their solutions depends on the value under the square root in the quadratic formula, called the discriminant. The discriminant is given by: \[ \Delta = b^2 - 4ac \]
The value of \( \Delta \) determines the type of solutions:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is one real solution.
• If \( \Delta < 0 \), there are no real solutions (instead, two complex solutions).

In our exercise, the discriminant is calculated as follows: \[ ( -9 )^2 - 4 \times 1 \times 2 \] We get: \[ 81 - 8 = 73 \] Since \( 73 > 0 \), we have two distinct real solutions. Hence, the solutions are: \[ x = \frac{9 + \sqrt{73}}{2} \] and \[ x = \frac{9 - \sqrt{73}}{2} \]

Clearing fractions

When solving equations with fractions, it is often useful to eliminate the fractions early on. This simplifies the equation and makes further solving easier. To clear fractions:

• Identify a common denominator for all the fractions in the equation.
• Multiply every term in the equation by this common denominator.

In our exercise: \[ \frac{3 x}{x-2} + \frac{1}{x} = 4 \] The common denominator is \( x(x-2) \). By multiplying every term by \( x(x-2) \), we eliminate the fractions: \[ \frac{3x}{x-2} \times x(x-2) + \frac{1}{x} \times x(x-2) = 4 \times x(x-2) \] Simplifying this gives: \[ 3x^2 + (x-2) = 4x^2 - 8x \] This step-by-step removal of fractions sets up the equation in a form where we can apply the quadratic formula.