Problem 20 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the quadratic equation by factoring. Check your solutions in the original equation. $$x^{2}+2 a x+a^{2}=0$$

Step-by-Step Solution

Answer

The solution to the quadratic equation \(x^{2}+2 a x+a^{2}=0\) by factoring is \(x = -a\).

Step 1: Formulating the quadratic equation

The given quadratic equation is \(x^{2}+2ax+a^{2}=0\). It's important to note that it takes the form of a perfect square trinomial.

Step 2: Factoring the equation

The given quadratic can be factored as \((x + a)^2 = 0\). This is because the given equation fits the pattern of a perfect square trinomial, which can be factored as \(a^2 + 2ab + b^2 = (a+b)^2\). Here, \(a = x\) and \(b = a\).

Step 3: Solving for x

Setting the factored equation to zero, we have, \((x + a)(x + a) = 0\). For this product to be zero, at least one factor has to be zero. Therefore, \(x+a = 0\). Solved, gives us \(x = -a\).

Step 4: Checking the solution in the original equation

Substitute the solution \(x = -a\) into the original equation, we have, \(x^{2}+2ax+a^{2} = (-a)^{2}+2a(-a)+(a^{2}) = a^{2} - 2a^{2} + a^{2} = 0\). This validates the solution.

Key Concepts

Perfect Square Trinomial

A perfect square trinomial is a special type of quadratic expression that can be factored into a binomial squared. This means it fits the pattern

• \(a^2 + 2ab + b^2 = (a+b)^2\)

In the original exercise, we started with the quadratic equation \(x^2 + 2ax + a^2 = 0\). Recognizing this as a perfect square trinomial is key.
We see that the equation can be rewritten in the form \((x + a)^2 = 0\), demonstrating its perfect square nature. These expressions are particularly nice to work with because they simplify solving quadratics, thanks to their predictable pattern. Identifying this pattern is crucial for factoring quadratics efficiently. By comparing each term in the equation to the standard form of the perfect square, we ascertain that \(x^2\) corresponds to \(a^2\), while \(2ax\) can be matched to \(2ab\), where both variables \(a\) and \(b\) equal \(x\) and \(a\), respectively.

Solving Quadratic Equations

Solving quadratic equations often involves the process of finding values for \(x\) that satisfy the equation. In our case, the equation is already in the factored form \((x+a)^2 = 0\).
This indicates a neat method for solving – set the factor equal to zero.

• This gives us \((x+a)(x+a) = 0\), which simplifies to \(x+a = 0\).
• Therefore, the solution is \(x = -a\).

The beauty of using this method with perfect square trinomials is that it efficiently leads us to the solution. This step doesn't just tell us where \(x\) lies in terms of \(a\), it also helps confirm that our original understanding of the form of the quadratic was correct. Recognizing when an expression is a perfect square can save time and simplify the solving process greatly.

Checking Solutions in Equations

Verifying your solution is a crucial part of solving any equation, and quadratic equations are no exception. After solving the quadratic equation \(x^2 + 2ax + a^2 = 0\) and finding \(x = -a\), the final step is to substitute this value back into the original equation to ensure accuracy.
When we replace \(x\) with \(-a\), we compute:

• \((-a)^2 + 2a(-a) + a^2 = a^2 - 2a^2 + a^2\)
• This simplifies to \(0\).

Since the left side equals zero, which matches the right side of the equation, our solution is verified. Checking your work may seem tedious, but it confirms that your solution doesn't just work under special circumstances; it ensures accuracy across the board. Remember, confidence in mathematics comes through practice and verification.