Problem 54 - Introductory Algebra for College Students

Question

Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$4 y^{2}+44 y+121=0$$

Step-by-Step Solution

Answer

The solution to the equation \(4y^2 + 44y + 121 = 0\) is \(y = -\frac{11}{2}\).

Step 1: Rewrite the Equation

Keep all terms on one side of the equation, set equal to zero: \[4y^2 + 44y + 121 = 0\]

Step 2: Factor the Equation

Factoring the quadratic equation, it's observed that it is a perfect square trinomial. A perfect square trinomial is a special form of trinomial that can be factored into the square of a binomial. The equation can be factored as follows: \[(2y + 11)^2 = 0\]

Step 3: Apply Zero-Product Property

The Zero-Product Property states that if a * b = 0, then a = 0 or b = 0. Hence, setting the equation above to equal zero, we get: \[2y + 11 = 0\]

Step 4: Solve for y

Solving the above equation for y, we have: \[y = -\frac{11}{2}\]

Step 5: Check the Solution

Substitute -11/2 back into the original equation and confirm if it holds true: \[4(-11/2)^2 + 44(-11/2) + 121 = 0 \leftrightarrow 121 - 242 + 121 = 0 \leftrightarrow 0 = 0\] The solution holds true, hence, y = -11/2 is the correct solution.

Key Concepts

Perfect Square Trinomial

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). When factoring these equations, one form you might encounter is a perfect square trinomial. This is a special type of quadratic where the equation can be expressed as the square of a binomial. In general, a perfect square trinomial takes the form of \( (ax + b)^2 = a^2x^2 + 2abx + b^2 \). This means that both the first and third terms are perfect squares themselves, and the middle term is twice the product of the bases of those two terms.

In the equation \( 4y^2 + 44y + 121 = 0 \), we can recognize it as a perfect square trinomial because it can be written as \( (2y + 11)^2 = 0 \). Here, \( 4y^2 \) is \( (2y)^2 \) and \( 121 \) is \( 11^2 \), while \( 44y \) is twice the product of \( 2y \) and \( 11 \). This recognition helps us simplify the process by factoring the trinomial directly into a binomial squared.

Zero-Product Property

Once we have successfully factored a quadratic equation into binomials, the Zero-Product Property allows us to solve for the variable. This property is a fundamental aspect of algebra, and it asserts that if the product of two factors is zero, then at least one of the factors must be zero. Mathematically, if \( a \times b = 0 \), then \( a = 0 \) or \( b = 0 \).

Applying the Zero-Product Property to our factored equation \( (2y + 11)^2 = 0 \), we obtain the simple linear equation \( 2y + 11 = 0 \). By solving this equation, we can find the value of \( y \) that satisfies the original quadratic equation. This straightforward approach is one of the advantages of factoring quadratic equations into simpler binomial terms.

Solving Quadratic Equations

Solving quadratic equations typically involves finding the values of the variable that satisfy the equation. This process can be approached in several ways, including factoring, completing the square, or using the quadratic formula. Each method has its own set of benefits depending upon the form of the quadratic equation.

In our exercise, after applying the Zero-Product Property, we solved the linear equation \( 2y + 11 = 0 \) by isolating \( y \). This results in \( y = -\frac{11}{2} \).

• Validation: Remember to validate your solution either by substituting back into the original equation or using a graphing utility to check the solution's correctness.
• Graphical Method: Graphing can offer a visual representation to confirm that the value \( y = -\frac{11}{2} \) is indeed the point where the graph crosses the x-axis.

Checking is an essential part of solving as it verifies that no errors were made during the algebraic manipulation.