Problem 34 - Precalculus

Question

Use the quadratic formula to solve each equation. In Exercises \(34-39,\) give two forms for each solution: an expression containing a radical and a calculator approximation rounded off to two decimal places. $$2 x^{2}+3 x-4=0$$

Step-by-Step Solution

Answer

Solutions: \(\frac{-3 \pm \sqrt{41}}{4}\), \(x \approx 0.85\) and \(x \approx -2.85\).

Step 1: Identify Quadratic Equation Coefficients

The given quadratic equation is \(2x^2 + 3x - 4 = 0\). Identify the coefficients: \(a = 2\), \(b = 3\), and \(c = -4\).

Step 2: Use the Quadratic Formula

Recall the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the identified coefficients into the formula: \(x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-4)}}{2(2)}\).

Step 3: Simplify the Discriminant

Calculate the discriminant \(b^2 - 4ac\): \(3^2 - 4(2)(-4) = 9 + 32 = 41\).

Step 4: Calculate the Solutions with Radicals

Substitute the discriminant back into the quadratic formula: \(x = \frac{-3 \pm \sqrt{41}}{4}\). Therefore, the solutions in radical form are: \(x = \frac{-3 + \sqrt{41}}{4}\) and \(x = \frac{-3 - \sqrt{41}}{4}\).

Step 5: Calculate Approximate Solutions

Use a calculator to approximate \(\sqrt{41} \approx 6.40\). Compute each solution: \(x_1 = \frac{-3 + 6.40}{4} \approx 0.85\) and \(x_2 = \frac{-3 - 6.40}{4} \approx -2.85\).

Step 6: Present Both Forms of Solutions

The solutions in radical form are \(x = \frac{-3 + \sqrt{41}}{4}\) and \(x = \frac{-3 - \sqrt{41}}{4}\). The approximate solutions are \(x \approx 0.85\) and \(x \approx -2.85\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is typically written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients. To find the values of \( x \), we use the quadratic formula:

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

This formula allows us to find the two possible solutions for \( x \) by considering the "+" and "-" signs in front of the square root. By plugging in the coefficients \( a \), \( b \), and \( c \) from the equation, we can easily calculate \( x \) without having to factor the quadratic expression directly.

Discriminant

The discriminant is a critical element of the quadratic formula, and it plays a significant role in determining the nature of the roots. It is found inside the square root in the quadratic formula and is given by:

• \( b^2 - 4ac \)

The value of the discriminant helps us understand the type of solutions a quadratic equation might have:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If it is negative, the solutions are complex or imaginary.

In our example, the discriminant is calculated as \( 41 \), which is positive, indicating two distinct real solutions.

Radical Solutions

Radical solutions represent the solutions of the quadratic equation in their simplest exact form, including the square root, or radical, component. After simplifying the discriminant, we substitute it back into the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{41}}{4} \)

These solutions, \( x = \frac{-3 + \sqrt{41}}{4} \) and \( x = \frac{-3 - \sqrt{41}}{4} \), are called radical solutions because they involve a radical expression (\( \sqrt{41} \)). Radical solutions provide an exact answer rather than an approximation, which can be useful in many mathematical contexts and applications.

Calculator Approximation

While radical solutions give an exact result, sometimes an approximate numerical solution is more practical and easier to interpret, especially for applications in the real world. This is where calculator approximations come into play. To do this, calculate the square root to a certain number of decimal places and then evaluate the entire expression.For our solved equation, the square root of 41 is approximately 6.40. Using this to find the solutions:

• \( x_1 = \frac{-3 + 6.40}{4} \approx 0.85 \)
• \( x_2 = \frac{-3 - 6.40}{4} \approx -2.85 \)

These are rounded to two decimal places, making them suitable for situations where a quick estimate is needed rather than exact precision. This method is especially useful when solving quadratic equations by hand or using a calculator without a symbolic computation feature.