Question

Determine whether you would use factoring, square roots, or completing the square to solve the equation. Explain your reasoning. Then solve the equation. \(x^2+12 x+36=0\)

Short answer

The most appropriate method to solve the equation is factoring because the equation is a perfect square trinomial. The root of the equation is \(x = -6\).

Step-by-step solution

Identifying the Method

Considering the given equation \(x^2+12x+36=0\), notice that it is in the standard form of a quadratic equation, and can be written as \((x+6)^2 =0\). This is a perfect square. Therefore, the most appropriate method to apply here would be factoring.

Solving the Equation

We can factorize the equation as \((x+6)(x+6)=0\). For a product of two numbers to be zero, at least one of them has to be zero. Thus, we set each factor equal to zero and solve for \(x\), that is \(x+6 = 0\). Solving this equation for \(x\), we get \(x = -6\).

Checking the Solution

Substitute \(x=-6\) into the original equation to determine if the left side equals the right side. Plugging in yields \(36 + \(-6\)^2+12×\(-6\) = 0\), which simplifies to \(0 = 0\). Since this equation is true, it indicates that the solution is correct.

Key concepts

Factoring Quadratic Equations

Factoring is a powerful algebraic technique used to simplify expressions and solve equations, especially quadratic equations of the form \(ax^2 + bx + c = 0\). The goal is to break down the quadratic equation into the product of binomials.

When a quadratic equation has real-number solutions, it can often be factored into the form \( (x+n)(x+m)=0\), where \(-n\) and \(-m\) are the solutions for \(x\). The process involves finding two numbers that multiply to give the constant term, \(c\), and add up to the middle coefficient, \(b\).

Example in Textbook Solution

In the provided problem, \(x^2+12x+36=0\), we identified the form \((x+6)^2=0\), revealing that the equation is a perfect square and could be factored easily. This is because the factors are identical, which makes the solving process more straightforward. By setting each factor equal to zero, we find that the solution is \(x=-6\).

Perfect Square Trinomials

A perfect square trinomial is a special type of quadratic equation that can be expressed as the square of a binomial. In other words, it takes the form \((ax+b)^2 = a^2x^2 + 2abx + b^2\). The three terms in the trinomial are the result of squaring each term in the binomial and the cross product.

Recognizing a perfect square trinomial is useful: it allows us to factorise the quadratic equation quickly. We look for a quadratic term (\(a^2x^2\)), a linear term (\(2abx\)), and a constant term (\(b^2\)), where \((2ab)\) is the middle coefficient of the standard quadratic form.

Identifying Perfect Squares

In the exercise, \(x^2+12x+36\) is recognized as a perfect square because \(12\) is twice the product of the square root of \(36\) (which is \((2)(6)\)). The equation fits the form \( (x+6)^2\). Spotting a perfect square trinomial simplifies the solving process to finding the binomial that, when squared, gives back the trinomial.

Completing the Square

Completing the square is a method used to solve any quadratic equation by converting it into a perfect square trinomial. This method is particularly helpful when the quadratic equation cannot be easily factored by inspection.

To complete the square, we start by arranging the quadratic in the form of \(ax^2+bx+c=0\) and work to get a perfect square trinomial from the first two terms. Here are typical steps involved:

• Move the constant term \(c\) to the right side of the equation.
• If \(a\) is not equal to 1, divide the entire equation by \(a\) to make the coefficient of \(x^2\) equal to 1.
• Add the square of half the coefficient of \(x\) to both sides of the equation.
• Factor the left side as the square of a binomial.
• Simplify the right side and solve for \(x\).

While completing the square was not necessary for the exercise's given equation, understanding this technique is crucial as it is the basis for the quadratic formula and can be applied to any quadratic equation, making it a versatile tool in algebra.