Problem 18 - 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. $$2+6 v=9 v^{2}$$

Step-by-Step Solution

Answer

The solutions are \( v = \frac{1 + \sqrt{3}}{3} \) and \( v = \frac{1 - \sqrt{3}}{3} \).

Step 1: Rearrange the Equation

First, let's rearrange the equation into the standard quadratic form, which is presented as \( ax^2 + bx + c = 0 \). Our equation is currently \( 2 + 6v = 9v^2 \). Subtract \( 9v^2 \) from both sides to get:\[ 0 = 9v^2 - 6v - 2 \]

Step 2: Identify the Coefficients

To use the quadratic formula, we need to identify the coefficients \( a \), \( b \), and \( c \) from the equation \( 9v^2 - 6v - 2 = 0 \). Here, \( a = 9 \), \( b = -6 \), and \( c = -2 \).

Step 3: Apply the Quadratic Formula

The quadratic formula is \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values of \( a \), \( b \), and \( c \) into the formula:\[ v = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot (-2)}}{2 \cdot 9} \]This simplifies to:\[ v = \frac{6 \pm \sqrt{36 + 72}}{18} \]

Step 4: Simplify Under the Square Root

Calculate the value under the square root:\[ 36 + 72 = 108 \]Thus the equation becomes:\[ v = \frac{6 \pm \sqrt{108}}{18} \]

Step 5: Simplify the Square Root

Simplify \( \sqrt{108} \). Notice that \( 108 = 36 \cdot 3 \), so\[ \sqrt{108} = \sqrt{36 \times 3} = \sqrt{36} \times \sqrt{3} = 6\sqrt{3} \].Substitute this back into the equation:\[ v = \frac{6 \pm 6\sqrt{3}}{18} \]

Step 6: Simplify the Fraction

Factor \( 6 \) out of the numerator:\[ v = \frac{6(1 \pm \sqrt{3})}{18} \].Divide both the numerator and the denominator by \( 6 \):\[ v = \frac{1 \pm \sqrt{3}}{3} \].

Step 7: Present the Solutions

The solutions to the equation \( 2 + 6v = 9v^2 \) are:\[ v = \frac{1 + \sqrt{3}}{3} \] and \[ v = \frac{1 - \sqrt{3}}{3} \].

Key Concepts

Quadratic Formula

Solving quadratic equations can sometimes be challenging, but the quadratic formula makes it much easier. The quadratic formula is:

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

This formula helps you find the values of \( v \) that make the equation true. The expression under the square root, \( b^2 - 4ac \), is called the discriminant. The discriminant can tell us about the nature of the roots:

• If positive, there are two distinct real roots.
• If zero, there is exactly one real root.
• If negative, the roots are complex and not real.

Using the quadratic formula is straightforward. Once you've rearranged your equations into standard form, simply plug in the coefficients.

Standard Quadratic Form

Before using the quadratic formula, it is crucial to convert any given quadratic equation into the standard quadratic form:

• \( ax^2 + bx + c = 0 \)

This makes it easier to identify the coefficients needed for the formula. In our example, the original equation was \( 2 + 6v = 9v^2 \). To rearrange this into the standard form, you move all terms to one side so the equation becomes \( 9v^2 - 6v - 2 = 0 \). Notice how from this form, you can clearly see the terms for \( a \), \( b \), and \( c \). Remember, the standard form is a foundation for solving with the quadratic formula.

Simplifying Square Roots

After identifying the coefficients and substituting them into the quadratic formula, you often encounter square roots, like \( \sqrt{108} \). Simplifying square roots is a key step. To simplify, break down the number under the square root into its prime factors. For \( 108 \):

• \( 108 = 36 \times 3 \)
• \( \sqrt{36} \times \sqrt{3} = 6\sqrt{3} \)

Understanding this process of simplification ensures you handle square roots effectively. This can lessen the complexity of your solutions and make the final steps easier to manage.

Identifying Coefficients

Identifying the coefficients in a quadratic equation is like picking out the building blocks for solving it. Once the equation is in the standard form \( ax^2 + bx + c = 0 \), you can see:

• \( a \) as the coefficient of \( x^2 \)
• \( b \) as the coefficient of \( x \)
• \( c \) as the constant term

In our revised equation \( 9v^2 - 6v - 2 = 0 \), you can easily identify \( a = 9 \), \( b = -6 \), and \( c = -2 \). Recognizing these values quickly is pivotal for using the quadratic formula efficiently. Practice with different equations will improve your ability to spot these key numbers.