Problem 6 - Algebra 2

Question

Solve each equation. Check your solutions. $$ \frac{4 n^{2}}{n^{2}-9}-\frac{2 n}{n+3}=\frac{3}{n-3} $$

Step-by-Step Solution

Answer

No real solutions exist.

Step 1: Identify the common denominator

The given equation is \( \frac{4 n^{2}}{n^{2}-9} - \frac{2 n}{n+3} = \frac{3}{n-3}\). The first step is to find a common denominator for all fractions. The denominators are \( n^2 - 9 = (n+3)(n-3) \), \( n+3 \), and \( n-3 \). Thus, the common denominator is \((n+3)(n-3)\).

Step 2: Rewrite the equation with the common denominator

Multiply each term by the common denominator \((n+3)(n-3)\) to eliminate the fractions: \( (n^2 - 9)\left(\frac{4n^2}{n^2-9}\right) - (n^2 - 9)\left(\frac{2n}{n+3}\right) = (n^2 - 9)\left(\frac{3}{n-3}\right) \)Simplifying, we get:\( 4n^2 (n+3)(n-3) - 2n(n-3) = 3(n+3) \)

Step 3: Simplify the equation

Distribute and simplify each term:1. \(4n^2(n-3) - 4n^2(n+3) = 4n^3 - 12n^2 - 4n^3 - 12n^2 = -24n^2\)2. \( -2n(n-3) = -2n^2 + 6n\)3. Combine the equations to get \(-24n^2 -2n^2 + 6n = 3n + 9\).

Step 4: Simplify further

Combine like terms: \(-26n^2 + 6n = 3n + 9\). Subtract \(3n\) from both sides:\(-26n^2 + 3n = 9\).

Step 5: Solve the quadratic equation

To solve \(-26n^2 + 3n = 9\), bring all terms to one side: \(-26n^2 + 3n - 9 = 0\). Use the quadratic formula:\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -26\), \(b = 3\), \(c = -9\).Calculate the discriminant:\(b^2 - 4ac = 3^2 - 4(-26)(-9) = 9 - 936 = -927\).Since the discriminant is negative, there are no real solutions.

Step 6: Conclusion

Since the discriminant is negative, this indicates there are no real solutions to the equation. It is impossible to find real values of \( n \) that satisfy the equation under the given conditions.

Key Concepts

Discriminant

When solving quadratic equations, understanding the discriminant is crucial. Consider a general quadratic equation of the form \( ax^2 + bx + c = 0 \). The discriminant is part of the quadratic formula and is given by \( b^2 - 4ac \).

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, the equation has exactly one real root (or a repeated root).
• If the discriminant is negative, like in our exercise example, the equation has no real roots but two complex roots.

In our exercise, we calculated the discriminant for \(-26n^2 + 3n - 9 = 0\). Upon finding \(b^2 - 4ac = -927\), it was clear that there are no real solutions, because the negative value indicates the roots are complex. This concept is important for predicting the nature of the roots without even solving the equation entirely.

Common Denominator

When dealing with equations involving fractions, finding a common denominator simplifies calculations considerably. In our original equation \( \frac{4n^2}{n^2-9} - \frac{2n}{n+3} = \frac{3}{n-3} \), the first step was to identify a common denominator for all fractions.The denominators were \( n^2 - 9 \), \( n + 3 \), and \( n - 3 \). By factoring \( n^2 - 9 \) as \( (n+3)(n-3) \), we quickly identified \( (n+3)(n-3) \) as the common denominator. This common denominator allowed us to eliminate the fractions by multiplying each term by it, leading to a simpler expression.This process is akin to having the same "language" or "currency" among all terms, allowing them to interact and transform seamlessly. It is a vital skill when manipulating algebraic equations, especially those involving complex fractional terms.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It is represented as:\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides a method to find the roots of any quadratic equation by substituting the coefficients \( a \), \( b \), and \( c \).In the exercise, after reducing the fractions and simplifying the equation, we were left with the quadratic equation \(-26n^2 + 3n - 9 = 0\). By substituting \( a = -26 \), \( b = 3 \), and \( c = -9 \) into the quadratic formula, we attempted to find its roots.The quadratic formula is a universal solution method because it can always provide the roots if they exist, whether they are real or complex. In our example, however, using the formula showed that there were no real solutions, as the discriminant was negative.