Problem 71 - 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. $$ 5 x^{2}-9 x+1=0 $$

Step-by-Step Solution

Answer

The solutions are 1.681 and 0.119.

Step 1: Identify the Quadratic Equation

The quadratic equation given is \(5x^2 - 9x + 1 = 0\). Our goal is to find the roots of this equation, rounded to the nearest one-thousandth.

Step 2: Apply the Quadratic Formula

To find the roots, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For the equation \(5x^2 - 9x + 1 = 0\), identify \(a = 5\), \(b = -9\), and \(c = 1\).

Step 3: Calculate the Discriminant

The discriminant \(D\) is calculated as follows: \[ D = b^2 - 4ac = (-9)^2 - 4 \times 5 \times 1 = 81 - 20 = 61 \].

Step 4: Find the Square Root of the Discriminant

Compute \(\sqrt{61}\) using a calculator for better precision. The approximate value is \(\sqrt{61} \approx 7.810\).

Step 5: Compute the Two Solutions

Now substitute these values back into the quadratic formula:- For the first solution: \[ x_1 = \frac{-(-9) + 7.810}{2 \times 5} = \frac{9 + 7.810}{10} = \frac{16.810}{10} = 1.681 \]- For the second solution: \[ x_2 = \frac{-(-9) - 7.810}{2 \times 5} = \frac{9 - 7.810}{10} = \frac{1.190}{10} = 0.119 \].

Step 6: Rounding the Solutions

The solutions are already rounded to the nearest one-thousandth, yielding:- \(x_1 = 1.681\)- \(x_2 = 0.119\).

Key Concepts

Quadratic Formula

The quadratic formula is one of the most effective tools for finding the roots of a quadratic equation. A quadratic equation is generally expressed as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.
To find the roots, you can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• \(x\) are the solutions or roots of the quadratic equation.
• \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation.

This formula provides two potential solutions because of the \(\pm\) symbol, which indicates there can be both a positive and a negative square root result.
The beauty of the quadratic formula lies in its ability to solve any quadratic equation, regardless of whether the solutions are real or complex.

Discriminant

The discriminant is an important part of the quadratic formula, serving as a reliable indicator of the number and type of roots an equation has. It is calculated as the expression \(b^2 - 4ac\), found under the square root in the quadratic formula.

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it equals zero, there is exactly one real root, a scenario often termed a repeated or double root.
• When the discriminant is negative, the equation has two complex roots.

By examining the discriminant, you gain initial insight into the nature of the solutions you're dealing with, whether they are real numbers or complex numbers.

Roots of Quadratic Equation

The roots of a quadratic equation are the solutions, or values of \(x\), that satisfy the equation. In mathematical terms, these are the values which, when substituted back into the original equation \(ax^2 + bx + c = 0\), will render it true.
These roots are determined using the quadratic formula, taking into account the calculation of the discriminant.

• For \(5x^2 - 9x + 1 = 0\), the roots were approximated to be \(x_1 = 1.681\) and \(x_2 = 0.119\).

Finding these roots can hence be seen as solving a puzzle where each piece, such as the coefficients and discriminant, plays a crucial role in achieving the final solution.

Solution Approximation

Solution approximation involves finding a close numerical estimate of a root, often to a desired level of precision. With quadratic equations, this is crucial for real-world applications where exact solutions might not be practical or possible due to the irrational nature of some roots.

• When we have irrational solutions involving square roots, like in \(\sqrt{61}\), a calculator can provide a decimal approximation, making it easier to interpret and use.
• In our example, \(\sqrt{61} \approx 7.810\), which helps approximate the roots as \(x_1 = 1.681\) and \(x_2 = 0.119\).

Rounding these solutions to the nearest one-thousandth ensures they are practical for applications while maintaining as much accuracy as possible.