Question

Find the real solutions, if any, of each equation. Use any method. $$ \frac{3 x}{x+2}+\frac{1}{x-1}=\frac{4-7 x}{x^{2}+x-2} $$

Short answer

The only real solution is \( x = \frac{1}{3} \).

Step-by-step solution

Understand the Equation

Rewrite the given equation: \ \( \frac{3x}{x+2} + \frac{1}{x-1} = \frac{4-7x}{x^2 + x - 2} \). Recognize that \( x^2 + x - 2 \) can be factored.

Factor the Denominator

The denominator of the right-hand side \( x^2 + x - 2 \) can be factored as \( (x+2)(x-1) \). This gives the revised equation: \ \( \frac{3x}{x+2} + \frac{1}{x-1} = \frac{4-7x}{(x+2)(x-1)} \)

Combine the Left Side

Find a common denominator for the fractions on the left side. The common denominator is \( (x+2)(x-1) \). Combine the fractions: \ \( \frac{3x(x-1) + 1(x+2)}{(x+2)(x-1)} \)

Simplify the Numerator

Expand and simplify the numerator: \ \( 3x(x-1) + 1(x+2) = 3x^2 - 3x + x + 2 = 3x^2 - 2x + 2 \)

Equate the Numerators

Now, set the numerators equal to each other: \ \( 3x^2 - 2x + 2 = 4 - 7x \)

Rearrange into a Standard Quadratic Form

Move all terms to one side to form a quadratic equation: \ \( 3x^2 - 2x + 2 - 4 + 7x = 0 \) which simplifies to \( 3x^2 + 5x - 2 = 0 \)

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 3, b = 5, \text{and} c = -2 \).

Calculate the Discriminant

Calculate the discriminant \( \Delta \ = b^2 - 4ac = 5^2 - 4(3)(-2) = 25 + 24 = 49 \)

Find the Roots

Using the quadratic formula: \ \( x = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6} \). This gives \( x = \frac{1}{3} \) and \( x = -2 \).

Verify the Solutions

Check the solutions, considering the original denominators \((x+2)\text{ and }(x-1)\): \ For \( x = \frac{1}{3} \) both denominators are non-zero. \ For \( x = -2 \), the denominator \( x + 2 \) becomes zero. Therefore, \( x = -2 \) is not a valid solution.

Key concepts

quadratic equations

A quadratic equation is any equation that can be written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the provided exercise, the quadratic equation derived is \(3x^2 + 5x - 2 = 0\). This form is essential because it allows the use of various solution methods, like factoring, the quadratic formula, or completing the square. Quadratic equations are foundational in algebra and appear in various applications, such as physics, engineering, and finance.

factoring polynomials

Factoring polynomials breaks down a polynomial into simpler polynomials (factors) that, when multiplied together, produce the original polynomial. In the provided solution, the denominator \(x^2 + x - 2\) is factored into \((x + 2)(x - 1)\). This step is crucial to simplify rational equations, allowing us to find a common denominator and combine fractions. Factoring also aids in solving quadratic equations by setting each factor to zero and solving for \(x\). For example, if \(x^2 - 3x + 2 = 0\), factoring to \((x-1)(x-2) = 0\) provides the solutions \(x = 1\) and \(x = 2\). Practice recognizing common factoring techniques like grouping, difference of squares, and quadratic trinomials to build your skills.

discriminant

The discriminant in a quadratic equation \(ax^2 + bx + c = 0\) is the expression \(b^2 - 4ac\). It determines the nature and number of the solutions of the quadratic equation. In the provided solution, the discriminant is calculated for \(3x^2 + 5x - 2 = 0\) as follows: \(b^2 - 4ac = 5^2 - 4(3)(-2) = 25 + 24 = 49\). A positive discriminant (greater than zero) indicates two distinct real solutions. If it equals zero, the equation has exactly one real solution. A negative discriminant means there are no real solutions but two complex ones. Calculating the discriminant is a quick way to understand how many and what type of solutions to expect without solving the entire equation.