Problem 33
Question
For the following problems, solve the equations, if possible. $$ y(y+9)^{2}=0 $$
Step-by-Step Solution
Verified Answer
Answer: The solutions to the given equation are \(y = 0\) and \(y = -9\).
1Step 1: Consider case where y = 0
In this case, since \(y = 0\), the equation becomes \(0*(0+9)^{2}=0\). This is always true, so \(y=0\) is a solution.
2Step 2: Consider case where (y + 9)² = 0
In this case, we need to find the values of \(y\) that make \((y+9)^{2}=0\). Since squares of real numbers are always non-negative, the only way for the square to be zero is if the expression inside the square is zero as well. Therefore, we need to find the values of \(y\) that make \(y+9 = 0\). We can rewrite this equation as:
$$
y = -9
$$
Thus, the other solution to the given equation is \(y = -9\).
3Step 3: List all solutions
To summarize the solutions to the equation, we have found two possible values of \(y\) that satisfy the given equation. They are \(y = 0\) and \(y = -9\).
Key Concepts
Zero Product PropertyFactoring QuadraticsQuadratic Solutions
Zero Product Property
The zero product property is a fundamental aspect of algebra that states if the product of two or more factors is zero, then at least one of the factors must be zero. In other words, if we have an equation like \(a \times b = 0\), either \(a=0\) or \(b=0\) or both.
This property is extremely useful when solving quadratic equations, as it allows us to break down the equation into simpler parts. When we're faced with an equation like the given \(y(y+9)^{2}=0\), we can immediately see it takes the form of \(ab=0\) where \(a = y\) and \(b = (y+9)^2\). By applying the zero product property, we can set each factor equal to zero and solve for the variable separately, which simplifies finding the solutions immensely.
This property is extremely useful when solving quadratic equations, as it allows us to break down the equation into simpler parts. When we're faced with an equation like the given \(y(y+9)^{2}=0\), we can immediately see it takes the form of \(ab=0\) where \(a = y\) and \(b = (y+9)^2\). By applying the zero product property, we can set each factor equal to zero and solve for the variable separately, which simplifies finding the solutions immensely.
Factoring Quadratics
Factoring quadratics is a method of breaking down a quadratic equation into simpler components or 'factors' that, when multiplied together, give back the original equation. This technique is closely related to the zero product property. The process of factoring involves finding two binomials that, when expanded, produce the original quadratic expression.
For example, if we had a quadratic in the form of \(x^2 + bx + c\), we would search for two numbers that not only multiply to \(c\) but also add up to \(b\). However, the given exercise doesn't require factoring a typical quadratic expression but rather understanding that the expression \(y+9\) squared is already a product of two identical binomials \(y+9\). In the given problem, the quadratic expression was already in a factored state, making it easier to apply the zero product property.
For example, if we had a quadratic in the form of \(x^2 + bx + c\), we would search for two numbers that not only multiply to \(c\) but also add up to \(b\). However, the given exercise doesn't require factoring a typical quadratic expression but rather understanding that the expression \(y+9\) squared is already a product of two identical binomials \(y+9\). In the given problem, the quadratic expression was already in a factored state, making it easier to apply the zero product property.
Quadratic Solutions
Finding the solutions or 'roots' of quadratic equations involves determining the values that satisfy the equation when set equal to zero. Quadratic equations typically take the form \(ax^2 + bx + c = 0\), and the solutions can be found either by factoring, completing the square, using the quadratic formula, or as demonstrated in the textbook exercise, applying the zero product property.
The given equation has a slightly different structure since it's already set to zero, and it consists of a variable \(y\) and a squared term \(y+9)^2\. Upon applying the zero product property, we realize that the solutions are the values that make each factor zero. In our case, we have two solutions: \(y=0\) and \(y=-9\). It's important to note that each solution represents a point where the graph of the quadratic equation would cross the y-axis. Quadratic equations can have a maximum of two real solutions, as shown in the exercise we are discussing.
The given equation has a slightly different structure since it's already set to zero, and it consists of a variable \(y\) and a squared term \(y+9)^2\. Upon applying the zero product property, we realize that the solutions are the values that make each factor zero. In our case, we have two solutions: \(y=0\) and \(y=-9\). It's important to note that each solution represents a point where the graph of the quadratic equation would cross the y-axis. Quadratic equations can have a maximum of two real solutions, as shown in the exercise we are discussing.
Other exercises in this chapter
Problem 33
For the following problems, solve the equations by completing the square. $$ y^{2}-3 y+10=0 $$
View solution Problem 33
For the following problems, solve each of the quadratic equations using the method of extraction of roots. $$ 3 x^{2}-27=0 $$
View solution Problem 33
For the following problems, use the zero-factor property to solve the equations. $$ 4 m=0 $$
View solution Problem 34
For the following problems, solve the equations. $$ a^{2}+2 a=-1 $$
View solution