Problem 36 - College Algebra and Calculus: An Applied Approach

Question

Use a calculator to solve the quadratic equation. (Round your answer to three decimal places.) $$ -0.005 x^{2}+0.101 x-0.193=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x1 = -19.340 \) and \( x2 = -9.740 \).

Step 1: Substitute the values into the formula

Substitute the values of \( a = -0.005 \), \( b = 0.101 \), and \( c = -0.193 \) into the quadratic formula \( x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \). Doing so gives: \( x = -\frac{0.101 \pm \sqrt{(0.101)^{2} - 4(-0.005)(-0.193)}}{2(-0.005)} \)

Step 2: Calculate the value under the square root

Calculate the value of \( (0.101)^{2} - 4(-0.005)(-0.193) \). The result is 0.000121.

Step 3: Calculate the two potential solutions

There are two potential solutions based on the plus and minus in the formula. Calculating these gives the two potential solutions as: \( x1 = -\frac{0.101 + \sqrt{0.000121}}{2(-0.005)} = -19.34 \), \( x2 = -\frac{0.101 - \sqrt{0.000121}}{2(-0.005)} = -9.74 \)

Step 4: Round the solutions

According to the problem statement, the solutions need to be rounded to three decimal places. So the solutions are: \( x1 = -19.340 \), \( x2 = -9.740 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool frequently employed to find the roots of a quadratic equation, which is any equation that can be rearranged into the form of \( ax^2 + bx + c = 0 \). The roots of the equation are the values of \(x\) that make the equation true.

The formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It is derived from the process of completing the square and provides a systematic way to calculate the roots. To use the formula, you identify the coefficients \(a\), \(b\), and \(c\) from your equation and plug them into the formula. After calculating the discriminant, \(b^2 - 4ac\), you can determine the nature of the roots. If the discriminant is positive, there are two real and distinct roots. If it is zero, there is one real root (also known as a repeated or double root). If the discriminant is negative, the equation has two complex roots.

In the given example, we had \( a = -0.005 \), \( b = 0.101 \), and \( c = -0.193 \). Substituting these values into the quadratic formula and performing the necessary calculations, we find two distinct solutions, which are the roots of the equation.

Completing the Square

Completing the square is another method for solving quadratic equations and also serves as the basis for deriving the quadratic formula. The goal of completing the square is to transform a quadratic equation into a perfect square trinomial, which can then be easily solved by taking the square root of both sides.

The key steps involve:

• Rearranging the equation to ensure that the coefficient of the \(x^2\) term is 1 (if it’s not, divide the whole equation by the coefficient).
• Moving the constant term to the other side of the equation.
• Adding the square of half the coefficient of \(x\) to both sides to create a perfect square on one side.

Once you have a perfect square, you can express the equation in the form \((x + d)^2 = e\) where \((x + d)\) is your completed square and \((e\) is the value on the other side of the equation. Taking the square root of both sides gives you \(x + d = \pm\sqrt{e}\), and you can solve for \(x\) from here.

Roots of Quadratic Equation

The roots of a quadratic equation are the values of \(x\) that satisfy the equation, meaning the points where the quadratic function intersects the \(x\)-axis on a graph. There are various methods to find them, with the most common being factoring, using the quadratic formula, or completing the square.

The number and type of roots depend on the discriminant, \(b^2 - 4ac\), of the quadratic formula.

• If \(b^2 - 4ac > 0\), there are two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root, which is also called a repeated or double root.
• If \(b^2 - 4ac < 0\), the roots are complex and come in a conjugate pair.

In the provided exercise, by applying the quadratic formula, we found the roots to be \( x1 = -19.340 \) and \( x2 = -9.740 \), rounded to three decimal places. These roots are the solutions to the equation \( -0.005x^2 + 0.101x - 0.193 = 0 \).