Problem 51
Question
Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$y(y+8)=16(y-1)$$
Step-by-Step Solution
Verified Answer
The solution is \( y = 4 \).
1Step 1: Rewrite the equation
Let's start by opening out the brackets and bringing everything onto one side of the equal sign to get a quadratic equation.\( y^2 + 8y = 16y - 16 \) simplifies to \( y^2 - 8y + 16 = 0 \)
2Step 2: Factoring the quadratic equation
The equation \( y^2 - 8y + 16 = 0 \) can be factored to \( (y-4)^2 = 0 \) because the equation is a perfect square trinomial.
3Step 3: Solving for y
Setting \( (y-4)^2 = 0 \) gives us \( y = 4 \).
4Step 4: Verification
Substitute \( y = 4 \) back into the original equation to verify. \( 4(4+8) = 16(4-1) \), which simplifies to \( 48 = 48 \), thus confirming that the solution is valid.
5Step 5: Graphical verification
Another way to verify the solution is by plotting \( y(y+8) \) and \( 16(y-1) \) in a graphing utility and looking for the x-intercept. The x-intercept will be equal to the y-value obtained by solving the equation algebraically. In this case, the graph of both equations will intersect at \( y = 4 \) demonstrating that the solution is correct.
Key Concepts
Quadratic EquationPerfect Square TrinomialGraphical Verification
Quadratic Equation
A quadratic equation is a polynomial equation of degree two. It typically takes the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. Quadratic equations form a parabola when graphed, which can open upwards or downwards depending on the sign of \( a \).
Understanding how to rewrite an equation into the standard quadratic form is essential in solving or factoring it. In this exercise, we start from the equation \( y(y+8) = 16(y-1) \). By expanding and rearranging it, we get \( y^2 - 8y + 16 = 0 \).
This transformation showcases a methodical approach to handle quadratic equations:
Understanding how to rewrite an equation into the standard quadratic form is essential in solving or factoring it. In this exercise, we start from the equation \( y(y+8) = 16(y-1) \). By expanding and rearranging it, we get \( y^2 - 8y + 16 = 0 \).
This transformation showcases a methodical approach to handle quadratic equations:
- Expand all expressions fully.
- Move all terms to one side of the equation.
- Rearrange them to match the quadratic form.
Perfect Square Trinomial
A perfect square trinomial is a special type of quadratic expression that can be factored into an identical binomial squared. It follows the pattern \( (x + b)^2 = x^2 + 2bx + b^2 \). Recognizing a perfect square trinomial allows for quick and efficient factoring.
In our case, the equation \( y^2 - 8y + 16 = 0 \) is a perfect square trinomial. This equation can be expressed as \( (y-4)^2 = 0 \). When the trinomial is factored into \( (y-4)^2 \), it shows us that the solution for \( y \) is found by solving \((y-4) = 0 \), which gives \( y = 4 \).
The beauty of perfect square trinomials lies in their symmetrical nature, making them straightforward to factor. Recognizing them allows you to not only solve the equation quickly but also to reduce potential mistakes in the calculation process.
In our case, the equation \( y^2 - 8y + 16 = 0 \) is a perfect square trinomial. This equation can be expressed as \( (y-4)^2 = 0 \). When the trinomial is factored into \( (y-4)^2 \), it shows us that the solution for \( y \) is found by solving \((y-4) = 0 \), which gives \( y = 4 \).
The beauty of perfect square trinomials lies in their symmetrical nature, making them straightforward to factor. Recognizing them allows you to not only solve the equation quickly but also to reduce potential mistakes in the calculation process.
Graphical Verification
Graphical verification is a valuable technique to confirm solutions obtained algebraically. This involves graphing the functions described by the equation and checking the visual intercepts or intersection points.
According to the original exercise, after solving \( y^2 - 8y + 16 = 0 \) and obtaining \( y = 4 \) algebraically, we verify graphically. By plotting both \( y(y+8) \) and \( 16(y-1) \) on a graphing utility, the intersection of their graphs provides confirmation of the solution.
This step assures that both algebraic and graphical solutions coincide, offering a deeper understanding of the problem. Graphical analysis is especially useful when dealing with complex problems where visualization provides insights. It acts as a double-check mechanism to safeguard against possible algebraic errors.
According to the original exercise, after solving \( y^2 - 8y + 16 = 0 \) and obtaining \( y = 4 \) algebraically, we verify graphically. By plotting both \( y(y+8) \) and \( 16(y-1) \) on a graphing utility, the intersection of their graphs provides confirmation of the solution.
This step assures that both algebraic and graphical solutions coincide, offering a deeper understanding of the problem. Graphical analysis is especially useful when dealing with complex problems where visualization provides insights. It acts as a double-check mechanism to safeguard against possible algebraic errors.
- Plot both expressions independently.
- Identify the intersection point.
- Ensure it matches the algebraic solution.
Other exercises in this chapter
Problem 51
Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations usin
View solution Problem 51
Factor completely. $$2 r^{3}+6 r^{2}+4 r$$
View solution Problem 51
Factor each polynomial using the negative of the greatest common factor. $$-8 x^{4}+32 x^{3}+16 x^{2}$$
View solution Problem 51
Factor any perfect square trinomials, or state that the polynomial is prime. $$x^{2}+22 x+121$$
View solution