Problem 20 - 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. $$62 x+63=40 x^{2}$$

Step-by-Step Solution

Answer

The solutions are \(x = 2.25\) and \(x = -0.7\).

Step 1: Rearrange the Equation

First, bring all terms to one side of the equation to set it to zero. The given equation is \(62x + 63 = 40x^2\). Rearrange it to: \(40x^2 - 62x - 63 = 0\).

Step 2: Identify Coefficients

Identify the coefficients to use with the quadratic formula \(ax^2 + bx + c = 0\). Here, \(a = 40\), \(b = -62\), and \(c = -63\).

Step 3: Write the Quadratic Formula

The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into the formula.

Step 4: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\). Calculating gives \((-62)^2 - 4 \times 40 \times (-63) = 3844 + 10080 = 13924\).

Step 5: Evaluate the Quadratic Formula

Substitute the values into the quadratic formula: \(x = \frac{62 \pm \sqrt{13924}}{80}\). Calculate \(\sqrt{13924}\) to proceed.

Step 6: Simplify the Expression

Find \(\sqrt{13924} = 118\), then substitute back: \(x = \frac{62 \pm 118}{80}\). This gives two potential solutions: \(x = \frac{180}{80}\) and \(x = \frac{-56}{80}\).

Step 7: Simplify Further

Simplify both fractions: \(x = \frac{180}{80} = 2.25\) and \(x = \frac{-56}{80} = -0.7\). The solutions to the quadratic equation are \(x = 2.25\) and \(x = -0.7\).

Key Concepts

Quadratic Formula

Quadratic equations are a crucial part of algebra and can often look intimidating at first glance. However, they can be solved in a straightforward manner using the quadratic formula. This formula provides a reliable way to find the solutions — or roots — of any quadratic equation given in the standard form:

• The standard form is: \(ax^2 + bx + c = 0\)

Here, \(a\), \(b\), and \(c\) are coefficients found in the equation.

The magical quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]You use this formula by substituting the coefficients \(a\), \(b\), and \(c\) from your specific equation into it.

The beauty of the quadratic formula lies in its ability to handle all sorts of quadratic equations, whether they produce real or complex solutions. Whenever you face a quadratic, remember this formula and you'll find the answer!

Discriminant

The discriminant is an integral part of the quadratic formula and is what makes it powerful in determining the nature of the roots. It is the expression under the square root in the formula:

• The discriminant is given as: \(b^2 - 4ac\)

This value tells us several important details about the quadratic roots:

• If \(b^2 - 4ac > 0\), the equation has two distinct real roots.
• If \(b^2 - 4ac = 0\), there is exactly one real root, also known as a repeated or double root.
• If \(b^2 - 4ac < 0\), the equation has two complex roots.

For the given equation \(62x + 63 = 40x^2\), we calculated the discriminant to be \(13924\), which is greater than zero, indicating two distinct real roots.

Understanding the discriminant is essential since it quickly provides information about the type and number of roots without solving the entire equation.

Coefficients

Coefficients are numbers that multiply the variables in an equation. They play a critical role in solving quadratic equations using the quadratic formula. For the formula \(ax^2 + bx + c = 0\):

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant, or the term without a variable

In our example problem, the rearranged equation \(40x^2 - 62x - 63 = 0\) had the coefficients:

• \(a = 40\)
• \(b = -62\)
• \(c = -63\)

Correct identification of these coefficients is essential for correctly applying the quadratic formula.

By substituting \(a\), \(b\), and \(c\) into the formula, you can calculate the discriminant and ultimately solve for the roots. This highlights how critical each coefficient's role is in unlocking the solutions to the equation.