Problem 16 - Elementary Algebra

Question

Solve each quadratic equation by extraction of roots. $$ (y+11)^{2}-9=0 $$

Step-by-Step Solution

Answer

Question: Solve the quadratic equation \((y+11)^2 - 9 = 0\) by extraction of roots. Answer: The solutions are \(y = -8\) and \(y = -14\).

Step 1: Rewrite the equation as a perfect square

Add 9 to both sides of the equation to make the left side a perfect square: $$ (y+11)^2 - 9 + 9 = 0 + 9 $$ $$ (y+11)^2 = 9 $$

Step 2: Apply the square root property

Take the square root of both sides of the equation to isolate y: $$ \sqrt{(y+11)^2} = \sqrt{9} $$ Note that when taking the square root of both sides, we need to consider both the positive and negative roots: $$ y + 11 = \pm\sqrt{9} $$

Step 3: Solve for y

Since \(\sqrt{9} = 3\), substitute the value and solve for y: $$ y + 11 = \pm 3 $$ For the positive solution: $$ y + 11 = 3 $$ $$ y = 3 - 11 $$ $$ y = -8 $$ For the negative solution: $$ y + 11 = -3 $$ $$ y = -3 - 11 $$ $$ y = -14 $$ So, the solutions are \(y = -8\) and \(y = -14\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and are identifiable by their highest degree being two. For example, an equation in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not zero, is known as a quadratic equation. These equations pop up across various disciplines and can have two, one, or no real solutions.

To solve a quadratic equation, one might use methods like factoring, completing the square, using the quadratic formula, or extracting roots, depending on the equation's form. In the context of our exercise, we're interested in a special case of the quadratic equation that is already a perfect square trinomial or can be easily converted to one, allowing for the application of the square root property.

Square Root Property

The square root property is essential when dealing with quadratic equations that can be rewritten as a perfect square. It states that if \( x^2 = k \), then \( x = \pm\sqrt{k} \), where \( k \) is a non-negative real number. This property gives us two solutions: one positive root and one negative root, due to the nature of squaring a number: squaring both positive and negative versions of a number yields the same result.

The utility of this property is showcased in our exercise: after converting the equation to the form \( (y+11)^2 = 9 \), we can apply the square root property. By taking the square roots of both sides, we find that \( y + 11 \) must equal either the positive or negative value of \( \sqrt{9} \), ultimately leading to our two solutions for \( y \).

Perfect Square Trinomials

Perfect square trinomials are algebraic expressions that result from squaring a binomial. When a binomial is squared, it creates three terms, hence 'trinomial.' A trinomial is a perfect square if it can be written in the form \((a+b)^2 = a^2 + 2ab + b^2\) or \((a-b)^2 = a^2 - 2ab + b^2\). Recognizing these patterns is crucial because it allows for the application of the square root property to solve the equation.

In our textbook exercise, \((y+11)^2\) is already a perfect square trinomial. Understanding that \((y+11)^2\) must equal \(y^2 + 22y + 121\) if expanded, makes it clear why adding 9 to both sides sets the stage for using the square root property. This ability to manipulate an equation to reveal a perfect square is a powerful tool for solving quadratic equations. Recognizing perfect square trinomials simplifies the solving process and is a key aspect of algebraic proficiency.