Problem 18 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations using the quadratic formula. If there are no real roots, state this as the answer. Exercises \(3-6\) are the same as Exercises \(13-16\) of Section 7.2. $$62 x+63=40 x^{2}$$

Step-by-Step Solution

Answer

The solutions are \(x = 2.3\) and \(x = -0.75\).

Step 1: Rearrange the Equation

The given equation is \(62x + 63 = 40x^2\). First, we need to rearrange this equation into the standard quadratic form \(ax^2 + bx + c = 0\). This can be done by subtracting \(62x\) and \(63\) from both sides, getting \(40x^2 - 62x - 63 = 0\).

Step 2: Identify Coefficients

Now, identify the coefficients in the quadratic equation \(40x^2 - 62x - 63 = 0\). Here, \(a = 40\), \(b = -62\), and \(c = -63\).

Step 3: Write the Quadratic Formula

Recall the quadratic formula for solving equations of the form \(ax^2 + bx + c = 0\). The formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\) to determine if there are real roots. Substitute the values: \((-62)^2 - 4(40)(-63)\). Calculate \(3844 + 10080 = 14824\). The discriminant is \(14824\), which is positive, indicating two real roots.

Step 5: Use the Quadratic Formula

Using the quadratic formula with the values \(a = 40\), \(b = -62\), and \(c = -63\), substitute into the formula: \[x = \frac{62 \pm \sqrt{14824}}{80}\].

Step 6: Simplify the Expression

First, find the square root of \(14824\), which is \(122\). Substitute the values: \[x = \frac{62 \pm 122}{80}\]. Compute the two possible values for \(x\):1. \(x_1 = \frac{62 + 122}{80} = \frac{184}{80} = 2.3\).2. \(x_2 = \frac{62 - 122}{80} = \frac{-60}{80} = -0.75\).

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is a square. These equations are typically written in the standard form: \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Understanding the structure of this standard form is key to solving quadratic equations effectively.

• The term \( ax^2 \) is called the quadratic term.
• \( bx \) is the linear term.
• \( c \) is the constant term.

To solve these equations, one common method is to apply the quadratic formula. By rearranging the given equation \( 62x + 63 = 40x^2 \) into \( 40x^2 - 62x - 63 = 0 \), you can clearly identify your coefficients: \( a = 40 \), \( b = -62 \), and \( c = -63 \). Remember, all quadratic equations can take forms where real, imaginary, or no solutions exist based on the discriminant.

Discriminant

The discriminant is an integral part of the quadratic formula and a crucial concept in determining the nature of the roots of a quadratic equation. Represented by \( \Delta \), it is calculated using the formula \( b^2 - 4ac \). This value tells us about the roots of the equation quite effectively.

• If \( \Delta > 0 \), the equation has two distinct real roots.
• If \( \Delta = 0 \), there is exactly one real root, also known as a repeated or double root.
• If \( \Delta < 0 \), there are no real roots, only complex or imaginary roots.

For the quadratic equation in our example, the discriminant is \( 14824 \). Calculated by \( (-62)^2 - 4(40)(-63) \), the positive result indicates two distinct real roots. This diagnostic calculation helps us decide the next steps in solving the equation efficiently.

Real Roots

Real roots are solutions to quadratic equations that are actual numbers on the real number line. When the discriminant is positive or zero, as in our example, real roots exist. Using the quadratic formula, \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], one can determine these solutions.

• The term \( \pm \) implies that there might be two solutions, derived from adding and subtracting the square root of the discriminant.
• If the discriminant is zero, both solutions are the same, providing a single point where the parabola touches the x-axis.
• For positive discriminant values, as calculated for \( b^2 - 4ac = 14824 \), find \( x_1 \) and \( x_2 \) by substituting the values back into the quadratic formula.

In our solved example, the calculated roots are \( 2.3 \) and \( -0.75 \), showing where the graph of the equation crosses the x-axis. This visualization helps in understanding why both solutions are essential depending on the application scenario.