Problem 72 - 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. $$ 2 x^{2}-11 x-5=0 $$

Step-by-Step Solution

Answer

The solutions are approximately \(-0.422\) and \(5.922\).

Step 1: Recognizing the Equation Form

The given equation is a quadratic: \(2x^2 - 11x - 5 = 0\). This equation is in the standard form \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = -11\), and \(c = -5\).

Step 2: Using the Quadratic Formula

To solve the equation, we use the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\).

Step 3: Calculating the Discriminant

First, calculate the discriminant \(b^2 - 4ac\). Here, \(b = -11\), \(a = 2\), \(c = -5\).\(\text{Discriminant} = (-11)^2 - 4 \cdot 2 \cdot (-5) = 121 + 40 = 161\)

Step 4: Finding the Square Root of the Discriminant

Compute the square root of the discriminant: \(\sqrt{161}\). With a calculator, this gives approximately 12.6886.

Step 5: Substituting Values into the Formula

Now substitute these values into the quadratic formula:\[x = \frac{{11 \pm 12.6886}}{4}\]

Step 6: Calculating Each Solution

Calculate each solution separately:1. For \(x = \frac{{11 + 12.6886}}{4}\), we find \(x \approx 5.922\).2. For \(x = \frac{{11 - 12.6886}}{4}\), we find \(x \approx -0.422\).

Step 7: Expressing the Solution Set

The solutions, rounded to the nearest one-thousandth, are \(x = 5.922\) and \(x = -0.422\). Therefore, the solution set is \(\{-0.422, 5.922\}\).

Key Concepts

Quadratic Formula

The quadratic formula gives an easy way to find solutions to quadratic equations. These are equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The formula is expressed as:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

It is a powerful tool for solving any quadratic equation as long as you know the coefficients \(a\), \(b\), and \(c\). When using this formula, you calculate the expression under the square root sign, called the discriminant. This helps determine if there are real solutions. Once you have the value for the discriminant, you substitute it back into the formula, calculate, and find your solutions. This formula will provide two values, called roots, because of the plus and minus sign in the equation.

Rational Approximation

Rational approximation comes into play especially when dealing with square roots. Some numbers aren't perfect squares, so their square roots are irrational. This means that the decimal form of the square root goes on forever without repeating. For practical purposes, such numbers need to be approximated.

Using a calculator, you can find an approximate decimal form for such square roots. This approximation is crucial, especially in real-world problems requiring precise measurements. Rational approximation simplifies the use of these numbers by rounding them to acceptable decimal places, making calculations easier and more manageable. For example, in the previous problem, the square root of 161 was approximated to 12.6886, allowing further calculation with the quadratic formula.

Discriminant Calculation

The discriminant is a key part of the quadratic formula. It's the part under the square root, \(b^2 - 4ac\). Its value tells us how many and what type of solutions a quadratic equation has:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution.
• If it is negative, there are no real solutions, but two complex ones.

In our example, we calculated the discriminant by substituting \(b = -11\), \(a = 2\), and \(c = -5\) into the formula. Doing this tells us that the discriminant is 161, a positive number. Therefore, the equation has two distinct real roots, facilitating further calculations using the quadratic formula to find the exact roots.

Rounding Numbers

Rounding numbers is a fundamental math skill used during approximation. It simplifies results to make them easier to understand and use. In the context of quadratic equations, it's essential when dealing with results like square roots or rational approximations.

To round a number to the nearest one-thousandth means to keep three decimal places and adjust based on the fourth. If the fourth digit beyond the decimal is 5 or more, the last preserved digit rounds up. For instance, in the solution set of \{-0.422, 5.922\}, each number is rounded to the nearest one-thousandth.

• For \(5.922\), the number already had three digits, so no change was needed after calculation.
• For \(-0.422\), it was rounded from \(-0.42165...\) using the four-digit rule.

Rounding helps contain numbers within a practical framework, especially essential in applications requiring precision.