Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Use the Quadratic Formula to solve the quadratic equation. $$ 25 h^{2}+80 h+61=0 $$

Step-by-Step Solution

Answer

The solutions of the quadratic equation are \(h = -1.2536\) and \(h = -1.9464\).

Step 1: Identify Coefficients

From the given quadratic equation \(25h^2 + 80h + 61 = 0\), we can see that \(a = 25\), \(b = 80\), and \(c = 61\).

Step 2: Calculate the Discriminant

Next, calculate the value of the discriminant, which is \(b^2 - 4ac\). So discriminant \(D = 80^2 - 4*25*61 = 6400 - 6100 = 300\).

Step 3: Apply the Quadratic Formula

Now, apply these values into the quadratic formula \(h = \frac{-b \pm \sqrt{D}}{2a}\). Putting the values, we get \(h = \frac{-80 \pm \sqrt{300}}{50}\), which simplifies to \(h = -1.6 \pm 0.3464\).

Step 4: Solve for h

Finally, solve for both values of \(h\). From here, we find the two possible values: \(h = -1.2536\) and \(h = -1.9464\).

Key Concepts

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable with the standard form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \), are coefficients and \( a \) is not equal to zero. If \( a \) was zero, the equation would not be quadratic but linear. The solutions to these equations are the values of the variable that satisfy the equality.

The equation \( 25h^2 + 80h + 61 = 0 \) is a classic example of a quadratic equation, where \( h \) is the variable, and 25, 80, and 61 are the coefficients \( a \) (25), \( b \) (80), and \( c \) (61), respectively. The solutions to a quadratic equation can be real or complex and are found using methods like factoring, completing the square, or most commonly using the quadratic formula, which is applicable in all cases.

Discriminant Calculation

The discriminant is a crucial component in the quadratic formula; it determines the nature and number of solutions to a quadratic equation. The discriminant is represented by \( D \) and is calculated with the formula \( D = b^2 - 4ac \).

In the context of our example \( 25h^2 + 80h + 61 = 0 \), we identify \( a = 25 \) and \( b = 80 \) and \( c = 61 \) from the given quadratic equation, then we calculate the discriminant as follows:

• First square \( b \): \( 80^2 = 6400 \)
• Multiply \( 4 \), \( a \) and \( c \): \( 4*25*61 = 6100 \)
• Subtract the product from the squared \( b \) to find \( D \) : \( 6400 - 6100 = 300 \)

A positive discriminant, such as 300 in our exercise, indicates that there are two distinct real number solutions to the quadratic equation.

Algebraic Solutions

Algebraic solutions are fundamentally the answers we find when solving algebraic equations, such as a quadratic equation. Upon obtaining the discriminant, we proceed with the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \) to determine the algebraic solutions.

For our example, with \( a = 25 \) , \( b = 80 \) and the discriminant \( D = 300 \) ascertained, we apply these to the quadratic formula:

• Negative \( b \) and square root of \( D \) are inserted into the formula: \( h = \frac{-80 \pm \sqrt{300}}{2 \cdot 25} \)
• This simplifies to \( h = -1.6 \pm 0.3464 \) (by performing the operations, dividing \( -b \) by \( 2a \) and taking the square root of \( D \) )

Next, we calculate the two possible algebraic solutions, which include both the addition and subtraction of the square root result from the equation. In our exercise, this leads to the solutions for \( h \): \( h = -1.2536 \) and \( h = -1.9464 \). These are the points at which the quadratic equation equals zero, graphically representing as the points where the parabola (graphic representation of a quadratic equation) intersects with the horizontal axis.