Problem 70 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ h^{2}+2 h+1=0 $$

Step-by-Step Solution

Answer

The solution is \(h = -1\).

Step 1: Recognize the equation form

The given equation is a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = 2\), and \(c = 1\). It looks like a perfect square quadratic equation.

Step 2: Identify Perfect Square

Notice that the equation \(h^2 + 2h + 1 = 0\) can be expanded into \((h + 1)^2 = 0\). This hints that the quadratic is a perfect square trinomial.

Step 3: Confirm the Perfect Square

Verify that the equation forms a perfect square by writing it as \((h+1)^2 = 0\). This involves using the formula \((a+b)^2 = a^2 + 2ab + b^2\), where \(a=h\) and \(b=1\).

Step 4: Solve for h

Set the perfect square equation equal to zero: \((h+1)^2 = 0\). Solve for \(h\) by finding the value that makes the expression zero, which is \(h + 1 = 0\).

Step 5: Simplify the Solution

Solving \(h + 1 = 0\) gives \(h = -1\). This is the solution to the original quadratic equation.

Key Concepts

Understanding Perfect Square Trinomials

The concept of perfect square trinomials is central to solving certain types of quadratic equations effortlessly. A perfect square trinomial arises when you have an expression that can be expressed as the square of a binomial. In simple terms, it looks like this:

• If you have \(h^2 + 2h + 1\), it can be rewritten as \((h + 1)^2\).
• This follows from the formula \((a + b)^2 = a^2 + 2ab + b^2\). The first term \(h^2\) is \(a^2\), the middle term, \(2h\), is \(2ab\), and the last term \(1\) is \(b^2\).

Understanding this transformation is critical because it simplifies solving the equation. Instead of dealing with a quadratic trinomial, you're working with a squared binomial, making it easier to solve by taking the square root on both sides. Once identified, solving a perfect square trinomial becomes a straightforward task of setting the binomial equal to zero.

Breaking Down Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\). They represent curves known as parabolas when plotted on a graph. Let's break it down:

• The term \(ax^2\) is quadratic because it involves \(x\) squared. The coefficient \(a\) determines the parabola's width and direction.
• The linear component \(bx\) influences the parabola's axis and tilt.
• Finally, \(c\) represents the constant term, shifting the graph up and down.

To solve quadratic equations like \(h^2 + 2h + 1 = 0\), you'll often factor them, complete the square (as done for perfect square trinomials), or apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Each method has its use based on the equation's characteristics. Recognizing the type of quadratic equation you're dealing with is key to choosing the most straightforward solving strategy.

Step-by-Step Problem Solving Approach

A systematic, step-by-step approach is fundamental when solving quadratic equations. Here is a reliable method involving several straightforward steps:

• Identify the Equation: Recognize it as a quadratic equation. In our exercise, \(h^2 + 2h + 1 = 0\) is already in the standard quadratic form.
• Look for Patterns: Determine if it's a perfect square trinomial or suitable for factoring. In our example, the trinomial perfectly fits the identity \((h+1)^2\).
• Transform and Simplify: Rewrite the equation like \((h+1)^2 = 0\). This reduces complexity drastically.
• Solve the Transformed Equation: Equate the binomial to zero and solve for the variable \(h\). Here, you get \(h = -1\).

Using a step-by-step method ensures clarity, minimizing errors and helping you apply similar techniques to solve other quadratic equations efficiently. This structured approach is crucial, especially in more complex situations where direct observation might not quickly offer a solution.