Problem 65 - Intermediate Algebra

Question

The solution set for \(x^{2}-4 x-37=0\) is \(\\{2 \pm \sqrt{41}\\}\). With a calculator, we found a rational approximation, to the nearest one-thousandth, for each of these solutions. $$ 2-\sqrt{41}=-4.403 \quad \text { and } \quad 2+\sqrt{41}=8.403 $$ Thus the solution set is \(\\{-4.403,8.403\\}\), with the answers rounded to the nearest one-thousandth. Solve each of the equations in Problems 64-73, expressing solutions to the nearest one-thousandth. $$ x^{2}-16 x-24=0 $$

Step-by-Step Solution

Answer

The solutions are approximately \(-1.381\) and \(17.381\).

Step 1: Identify the Equation and Constants

We need to solve the equation \(x^2 - 16x - 24 = 0\) and express the solutions to the nearest one-thousandth. The equation is in standard quadratic form \(ax^2 + bx + c = 0\) with \(a = 1\), \(b = -16\), and \(c = -24\).

Step 2: Apply the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For this equation, substitute \(a = 1\), \(b = -16\), and \(c = -24\).

Step 3: Calculate the Discriminant

The discriminant of a quadratic equation is \(b^2 - 4ac\). Calculate it: \((–16)^2 - 4(1)(-24) = 256 + 96 = 352\).

Step 4: Solve for the Roots

Substitute the discriminant and other values into the quadratic formula:\[x = \frac{-(-16) \pm \sqrt{352}}{2(1)} = \frac{16 \pm \sqrt{352}}{2}\].

Step 5: Simplify and Find Approximations

To find \(\sqrt{352}\), we approximate it using a calculator, which gives approximately 18.761. Thus, simplify:\[x = \frac{16 \pm 18.761}{2}\].

Step 6: Calculate Each Solution

Calculate the two possible solutions:1. \(x = \frac{16 + 18.761}{2} \approx \frac{34.761}{2} = 17.3805\).2. \(x = \frac{16 - 18.761}{2} \approx \frac{-2.761}{2} = -1.3805\).

Step 7: Round the Solutions

Round each solution to the nearest one-thousandth:\(x \approx 17.381\) and \(x \approx -1.381\).

Step 8: State the Solution Set

Thus, the rounded solution set to the equation \(x^2 - 16x - 24 = 0\) is \(-1.381\) and \(17.381\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the solutions of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula can be applied to any quadratic equation and is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• \(a\), \(b\), and \(c\) are constants from the equation.
• \(\pm\) indicates that there will generally be two solutions.
• The formula uses the discriminant \(b^2 - 4ac\) to determine the number and type of solutions.

The quadratic formula is particularly useful when factoring is difficult or impossible. It provides a precise solution and helps in finding complex roots by setting up a framework to visualize all possibilities in a consistent manner.

Rational Approximation

Often in mathematics, the exact solutions to an equation can be irrational numbers. Enter rational approximations, which are estimates of these numbers in a more digestible form, often rounded to a certain number of decimal places. This is especially useful when dealing with square roots that cannot be simplified nicely.

• Rational approximation involves using a calculator to find a decimal approximation of an irrational root.
• This process makes it easier to understand and work with values, especially in real-world applications.

The challenge comes in using a rational approximation while maintaining the integrity of the solution, which demands accuracy. In our exercise, approximation plays a critical role, as it allows us to express the roots \(2 \pm \sqrt{41}\) as \(-4.403\) and \(8.403\) to the nearest thousandth.

Discriminant

The discriminant in a quadratic equation gives crucial insights into the nature of the roots of the equation. It is the part of the quadratic formula under the square root, denoted as \(b^2 - 4ac\).

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• A discriminant of zero indicates there is exactly one real root (also known as a repeated or double root).
• A negative discriminant means the equation has two complex roots.

In the given problem, the discriminant \(b^2 - 4ac\) equals \(352\), indicating two distinct real roots. This helps in understanding why two different solutions, \(-1.381\) and \(17.381\), were calculated.

Solution Set

A solution set in the context of quadratic equations is the set of values that satisfies the equation. After finding the solutions using the quadratic formula, we can express them as a set.With quadratic equations, the solution set typically includes two values due to the \(\pm\) in the formula. They are derived from calculating both \(+\) and \(-\) values of the expression:

• The solution \(x = \frac{16 + 18.761}{2}\) gives approximately \(17.381\) when rounded.
• The solution \(x = \frac{16 - 18.761}{2}\) gives approximately \(-1.381\) when rounded.

Therefore, the solution set \(\{ -1.381, 17.381 \}\) represents the values of \(x\) that satisfy the equation \(x^2 - 16x - 24 = 0\). These represent the points where the graph of the quadratic crosses the \(x\)-axis.