Problem 53 - 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. $$4 y^{2}+20 y+25=0$$

Step-by-Step Solution

Answer

By factoring the equation, we get \((2y+5)^2=0\). Solving for \(y\) gives \(y = -5/2\). Substituting this value in the original equation verifies the solution: inserting \(y = -5/2\) does indeed make the original equation true, confirming that the solution is valid.

Step 1: Factor the given equation

Start by factoring the quadratic equation \(4y^2+20y+25=0\). A quadratic equation in the form \(ax^2+bx+c=0\) can be factored into the form \((x-p)(x-q)=0\), where \(p\) and \(q\) are the roots of a quadratic equation. This is a special case called the perfect square trinomial where the form is \((a^2 + 2ab + b^2)\) which could be factored into \((a+b)^2\). Here, \(4y^2 = (2y)^2\), \(20y = 2*2y*5\), and 25 is \(5^2\). Hence, the equation can be factored as \((2y+5)^2\).

Step 2: Solving the factored equation

The factored form of our equation is \((2y+5)^2=0\). In this form, it is easy to find the solution by setting the factor equal to zero and solve for \(y\): \(2y+5 = 0\). Solving this equation will give the value of \(y\).

Step 3: Verify the solution

Once the solution is found, it should always be checked by substituting back into the original equation to ensure it satisfies the equation.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( x \) represents an unknown variable. These types of equations are called "quadratic" because the highest power of the variable is squared, or two. Quadratic equations can have either real or complex roots depending on the values of the coefficients. To identify a quadratic equation, simply check for the presence of an \( x^2 \) term, which makes the equation quadratic. The solutions to quadratic equations are essential in many areas of mathematics because they often describe real-world phenomena. Understanding how to solve and manipulate these equations is crucial for progressing in algebra and beyond.

There are various methods for solving quadratic equations, including factoring, using the quadratic formula, and completing the square. Each method has its advantages and can be applied depending on the specific form of the equation.

Perfect Square Trinomial

A perfect square trinomial is a special form of a quadratic expression that can be written as the square of a binomial. It takes the form \((a+b)^2\), which expands to \(a^2 + 2ab + b^2\). This pattern is key for recognizing when you can factor a quadratic equation as a binomial squared.

In the given exercise, the quadratic expression \(4y^2 + 20y + 25\) can be identified as a perfect square trinomial. By observing the terms, we see that \(4y^2 = (2y)^2\) and \(25 = 5^2\), with the middle term \(20y\) being twice the product of the square roots of these quantities. This makes it possible to rewrite the expression as \((2y + 5)^2\), making it easier to solve.

Solving Equations

Solving quadratic equations involves finding the values of the variable that make the equation true. Once you have factored a quadratic equation into its simplest form, such as a perfect square trinomial, you can set each factor equal to zero to find the solutions. In our step, we solve \((2y+5)^2=0\) by setting the factor \(2y+5=0\).

Setting the equation \(2y+5=0\) gives:
1. Subtract 5 from both sides to get \(2y = -5\).
2. Divide both sides by 2 to solve for \(y\), resulting in \(y = -\frac{5}{2}\).
Solving equations this way is straightforward when you recognize the equation is a perfect square trinomial, as it simplifies into a linear equation that can be managed easily.

Verification by Substitution

Verification by substitution is an essential step to ensure the solution obtained is correct. After solving a quadratic equation, substitute the solution back into the original equation to check if it satisfies the equation. This provides reassurance that no mistakes were made during solving.

In this case, our solution, \(y = -\frac{5}{2}\), is substituted back into the original quadratic equation:
1. Substitute \(y = -\frac{5}{2}\) into \(4y^2 + 20y + 25\).
2. Compute:
- \(4\left(-\frac{5}{2}\right)^2 = 4\times \frac{25}{4} = 25\)
- \(20\times -\frac{5}{2} = -50\)
- So, \(25 - 50 + 25 = 0 \)

This confirms that \(y = -\frac{5}{2}\) is indeed a valid solution because it satisfies the original equation.