Problem 41 - 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. $$24 x^{2}+23 x=-5$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{-1}{3}\) (\(\approx -0.33\)) and \(x = \frac{-5}{8}\) (\(\approx -0.63\)).

Step 1: Write the equation in standard form

Start by rewriting the given equation in standard form: \[ 24x^2 + 23x + 5 = 0 \] This is a requirement before applying the quadratic formula.

Step 2: Identify coefficients

Identify the coefficients from the standard form equation:\[ a = 24, \ b = 23, \ c = 5 \] These coefficients will be used in the quadratic formula.

Step 3: Apply the quadratic formula

Use the quadratic formula given by:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]Substitute the coefficients:\[ x = \frac{{-23 \pm \sqrt{{23^2 - 4 \times 24 \times 5}}}}{2 \times 24} \] Simplify within the square root first.

Step 4: Calculate the discriminant

Calculate the discriminant, \(b^2 - 4ac\):\[ b^2 - 4ac = 23^2 - 4 \times 24 \times 5 = 529 - 480 = 49 \]Since the discriminant is positive, there are two distinct real solutions.

Step 5: Simplify the expression with a radical

With the discriminant calculated, substitute back into the quadratic formula:\[ x = \frac{{-23 \pm \sqrt{49}}}{48} \]Since \(\sqrt{49} = 7\), the expression becomes:\[ x = \frac{{-23 \pm 7}}{48} \]

Step 6: Calculate exact radical form solutions

Solve for the two solutions using the expression:For the plus sign:\[ x = \frac{{-23 + 7}}{48} = \frac{-16}{48} = \frac{-1}{3} \]For the minus sign:\[ x = \frac{{-23 - 7}}{48} = \frac{-30}{48} = \frac{-5}{8} \]Thus, the solutions are \(\frac{-1}{3}\) and \(\frac{-5}{8}\).

Step 7: Find calculator approximations

Calculate the decimal approximations for each solution:For \(\frac{-1}{3}\):\[ x \approx -0.33 \]For \(\frac{-5}{8}\):\[ x \approx -0.63 \]These are the approximations rounded to two decimal places.

Key Concepts

Discriminant

The discriminant is a key part of the quadratic formula, and it helps us understand the nature of the solutions to a quadratic equation. It is found under the square root in the quadratic formula: \( b^2 - 4ac \).
Here's why it matters:

• If the discriminant is positive, there are two distinct real solutions, just as in our original exercise example.
• If the discriminant is zero, there's exactly one real solution; the parabola touches the x-axis at one point.
• If the discriminant is negative, there are no real solutions, only complex ones, because you can't take the square root of a negative number with real numbers.

Calculating the discriminant is an essential first step in analyzing any quadratic equation to predict the type of solutions you will encounter.

Real Solutions

Real solutions to a quadratic equation are the values of \(x\) that can be plotted on the real number line. Ensuring the discriminant is adequate is necessary for determining the existence of real solutions.
Once the discriminant is shown to be non-negative, the quadratic formula \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) can be fully evaluated. As seen in the example, our positive discriminant of 49 confirms there are two real solutions.
These solutions connect the x-intercepts of the parabola described by the quadratic equation. In graph terms, it's where the curve crosses or touches the x-axis.
Providing both the exact solutions in radical form and their decimal approximations gives a comprehensive understanding of the roots. It serves both precise calculations and practical approximations for applied situations.

Standard Form Equation

A quadratic equation in standard form looks like \( ax^2 + bx + c = 0 \). Before applying the quadratic formula, rewriting any quadratic equation into this form is crucial.
The coefficients \(a\), \(b\), and \(c\) pin down the specific characteristics of the equation:

• \(a\) dictates the direction and steepness of the parabolic curve (concave up if positive, concave down if negative).
• \(b\) skews the axis of symmetry of the parabola.
• \(c\) represents the y-intercept, where the parabola crosses the y-axis.

In our exercise, the original equation \(24x^2 + 23x = -5\) was restructured into standard form as \(24x^2 + 23x + 5 = 0\). This step lays the groundwork for effective use of strategies like the quadratic formula and confirms that none of the crucial elements are overlooked.