Problem 14 - Basic Technical Mathematics with Calculus

Question

solve the given quadratic equations, using the quadratic formula. Exercises \(5-8\) are the same as Exercises \(11-14\) of Section 7.2. $$4 x^{2}+x=3$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{3}{4} \) and \( x = -1 \).

Step 1: Write the Equation in Standard Form

For any quadratic equation, we must first arrange it in the standard form, which is \[ ax^2 + bx + c = 0 \].The given equation is:\[ 4x^2 + x = 3 \]Subtract 3 from both sides to set the equation to 0:\[ 4x^2 + x - 3 = 0 \].Thus, we have the equation in standard form where \( a = 4 \), \( b = 1 \), and \( c = -3 \).

Step 2: Apply the Quadratic Formula

The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Plug \( a = 4 \), \( b = 1 \), and \( c = -3 \) into the formula:\[ x = \frac{-(1) \pm \sqrt{(1)^2 - 4(4)(-3)}}{2(4)} \].

Step 3: Simplify the Expression Under the Square Root

Calculate the discriminant:\[ b^2 - 4ac = 1^2 - 4 \cdot 4 \cdot (-3) \]\[ = 1 + 48 \]\[ = 49 \].

Step 4: Compute the Roots Using the Quadratic Formula

Simplify further to find the roots:\[ x = \frac{-1 \pm \sqrt{49}}{8} \]Since \( \sqrt{49} = 7 \), we have:\[ x = \frac{-1 \pm 7}{8} \].

Step 5: Calculate Each Solution

First, solve for \( x \) with the positive sign:\[ x = \frac{-1 + 7}{8} = \frac{6}{8} = \frac{3}{4} \].Now, solve for \( x \) with the negative sign:\[ x = \frac{-1 - 7}{8} = \frac{-8}{8} = -1 \].

Key Concepts

Quadratic Formula

The quadratic formula is one of the most useful tools in algebra for finding the roots of quadratic equations. It provides an exact solution for any quadratic equation that is in the standard form. The formula is expressed as:

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

This formula allows us to solve any quadratic equation by simply substituting the values of coefficients \( a \), \( b \), and \( c \), which are derived from the equation's standard form. To effectively use the quadratic formula, it is crucial to understand how to manipulate and simplify the numbers within the formula, especially the expression under the square root, known as the discriminant. The double '±' sign signifies that two potential solutions (or roots) exist for the equation, which we will explore further in the upcoming sections.

Standard Form of Quadratic Equation

To solve a quadratic equation using the quadratic formula, the equation must first be expressed in its standard form. The standard form of a quadratic equation is:

• \( ax^2 + bx + c = 0 \)

Here, \( a \), \( b \), and \( c \) are known as the coefficients with \( a eq 0 \). The standard form ensures that the equation is set to zero, making it easier to identify the coefficients directly and substitute them into the quadratic formula. For instance, from the equation \(4x^2 + x = 3\), rearranging it to \(4x^2 + x - 3 = 0\) puts it in standard form with \(a = 4\), \(b = 1\), and \(c = -3\). This step is critical before applying the quadratic formula to solve for \(x\).

Discriminant

The discriminant is the part of the quadratic formula situated under the square root sign, represented by:

• \( b^2 - 4ac \)

This term is significant as it provides insight into the nature of the roots of the quadratic equation. The discriminant allows us to determine if the roots are real or complex without necessarily solving the equation. Depending on its value:

• If the discriminant is positive (), the equation has two distinct real roots.
• If it equals zero (), there is exactly one real root, meaning the roots are identical.
• If it is negative (), the equation has two complex conjugate roots.

In the given problem, the discriminant was calculated as \(49\), which is positive, indicating two distinct real roots. This step is often verified before proceeding to simplify for the actual values of the roots.

Roots of Quadratic Equations

The roots of a quadratic equation are the values for which the equation equals zero. These roots can be determined using the quadratic formula, where the term under the square root, namely the discriminant, plays a vital role. In the process outlined in our example, we determine the exact roots by solving:

• \( x = \frac{-1 \pm \sqrt{49}}{8} \)

Calculating further, we discover two solutions, \( x = \frac{3}{4} \) and \( x = -1 \). These values are known as the roots of the quadratic equation and represent the points where the parabola (the graph of the quadratic equation) intersects the x-axis. Understanding how to appropriately solve and interpret these roots can aid in graphing quadratic functions and analyzing their behavior.