Problem 70 - Algebra 2

Question

Solve each equation. Check your answers. \(\frac{2}{x+2}=\frac{6}{x^{2}-4}\)

Step-by-Step Solution

Answer

The solutions to the equation \(\frac{2}{x+2}=\frac{6}{x^{2}-4}\) are \(x=\sqrt{7}\) and \(x=-\sqrt{7}\).

Step 1: Simplify the equation

Rewrite \(x^{2}-4\) as \((x+2)(x-2)\) to simplify the equation. The equation becomes \(\frac{2}{x+2}=\frac{6}{(x+2)(x-2)}\).

Step 2: Multiply through by the common denominator

The common denominator in this equation is \((x+2)(x-2)\). Multiply both sides of the equation by the common denominator to get rid of fractions. After multiplication, you get \(2(x+2)(x-2)=6\).

Step 3: Simplify and solve for x

Expand the brackets on the left hand side to obtain \(2x^{2}-8=6\). Move 6 to the left-hand side and solve the quadratic equation \(2x^{2}-8-6=0\), which simplifies to \(2x^{2}-14 = 0\). Dividing through by 2, we have \(x^{2}-7=0\), giving us \(x^{2}=7\). After square rooting, we get two solutions: \(x=\sqrt{7}\) and \(x=-\sqrt{7}\).

Step 4: Check your solutions

When x equals \(\sqrt{7}\) or \(-\sqrt{7}\), the denominators \(x+2\) and \(x^{2}-4\) in original equation do not equal to 0, so they are valid. Substitute \(x=\sqrt{7}\) and \(x=-\sqrt{7}\) back into the original equation separately. After simplification, we find that both results are true, hence, both solutions are correct.

Step 5: State the solutions

The solution for the given equation \(\frac{2}{x+2}=\frac{6}{x^{2}-4}\) are \(x=\sqrt{7}\) and \(x=-\sqrt{7}\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). In our problem, after simplifying the rational equation and eliminating the denominators, we are left with the quadratic equation \(2x^2 - 8 = 6\). Here's how we solve it:
1. Subtract 6 from both sides to get all terms on one side: \(2x^2 - 14 = 0\).
2. Divide each term by 2 to simplify: \(x^2 -7 = 0\).
3. To find \(x\), take the square root of both sides, giving us two solutions: \(x = \sqrt{7}\) and \(x = -\sqrt{7}\).
Quadratic equations often have two solutions because squaring a negative and a positive number yields the same result. It's always crucial to consider both potential solutions when solving these types of equations.

Simplification

Simplification is an essential part of solving equations. It means making the equation easier to manage without changing its value. In this case, we simplify by rewriting \(x^2-4\) as \((x+2)(x-2)\). This factorization is possible because \(x^2-4\) is a difference of squares, which adheres to the formula \(a^2-b^2 = (a+b)(a-b)\).
This transformation allows us to see more clearly how the equation is structured and makes it easier to work with when multiplying through by the common denominator in the next steps. Simplification helps us to identify components that can be canceled or rearranged to make solving the equation straightforward.

Common Denominator

Finding a common denominator is a crucial step in solving equations involving fractions. To solve the rational equation \(\frac{2}{x+2} = \frac{6}{x^2-4}\), we need to eliminate the fractions by finding a common denominator, which is \((x+2)(x-2)\).
By multiplying both sides of the equation by this common denominator, we effectively "clear out" the fractions. This step transitions the equation from fractions to a simplified polynomial equation \(2(x-2) = 6\).
Removing fractions helps streamline the equation, making it more manageable and paving the way to applying algebraic techniques to find the solution.

Checking Solutions

After solving any equation, it's vital to check your solutions to ensure they work with the original equation. For rational equations, this involves substituting these solutions back into the original equation.
In our problem, the solutions found were \(x = \sqrt{7}\) and \(x = -\sqrt{7}\). Checking them involves substituting these values back into \(\frac{2}{x+2} = \frac{6}{x^2-4}\) and verifying that each side of the equation is equal for these \(x\) values.
Besides confirming equality, checking is also crucial to verify that the values do not lead to a zero denominator, as doing so would make them invalid. Thankfully, both solutions are valid as they do not result in a division by zero, and thus are correct.