Problem 20 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ 4 r^{2}-4 r-19=0 $$

Step-by-Step Solution

Answer

r = \frac{1}{2} + \sqrt{5} and r = \frac{1}{2} - \sqrt{5}

Step 1: Identify the coefficients

Given the quadratic equation \[ 4r^2 - 4r - 19 = 0 \],identify the values of the coefficients: a = 4, b = -4, and c = -19.

Step 2: Write the quadratic formula

Recall the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 3: Substitute the coefficients into the formula

Substitute the coefficients a = 4, b = -4, and c = -19 into the quadratic formula: \[ r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(-19)}}{2(4)} \]

Step 4: Simplify under the square root

Calculate the discriminant (the term under the square root): \[ (-4)^2 - 4(4)(-19) = 16 + 304 = 320 \]So, the equation becomes: \[ r = \frac{4 \pm \sqrt{320}}{8} \]

Step 5: Simplify the square root

Simplify \( \sqrt{320} \): \[ \sqrt{320} = \sqrt{64 \times 5} = 8 \sqrt{5} \]So, the equation becomes: \[ r = \frac{4 \pm 8 \sqrt{5}}{8} \]

Step 6: Simplify the fraction

Divide each term by 8: \[ r = \frac{4}{8} \pm \frac{8 \sqrt{5}}{8} = \frac{1}{2} \pm \sqrt{5} \]Thus, the solutions are: \[ r = \frac{1}{2} + \sqrt{5} \]and \[ r = \frac{1}{2} - \sqrt{5} \]

Key Concepts

Quadratic Equations

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations are called 'quadratic' because the highest power of the unknown variable \( x \) is 2. To solve quadratic equations, we can use various methods, including:

• Factoring
• Completing the square
• Using the quadratic formula

In the specific problem provided, we use the quadratic formula because it works for any quadratic equation, even if the other methods fail. The quadratic formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Understanding the parts of this formula is crucial. The expression inside the square root, \( b^2 - 4ac \), is called the discriminant.

Discriminant

The discriminant of a quadratic equation is the term \( b^2 - 4ac \). It plays an essential role in determining the nature of the solutions:

• If the discriminant is positive, the equation has two distinct real solutions.
• If the discriminant is zero, the equation has exactly one real solution (or a repeated root).
• If the discriminant is negative, the equation has no real solutions but two complex solutions.

In our exercise:
Given the quadratic equation \( 4r^2 - 4r - 19 = 0 \), we calculated the discriminant as follows:
\[ (-4)^2 - 4(4)(-19) = 16 + 304 = 320 \]Since the discriminant (320) is positive, we find that there are two distinct real solutions for \( r \).

Square Root Simplification

Simplifying square roots is important when solving quadratic equations using the quadratic formula. After substituting the coefficients into the formula, we often need to simplify the square root part, also known as the 'radical'. In our given problem, we first calculated the discriminant: \( \sqrt{320} \):
Since 320 can be factored into the product of 64 and 5, we can simplify it:
\[ \sqrt{320} = \sqrt{64 \times 5} = 8 \sqrt{5} \]Using this simplification makes it easier to deal with the rest of the quadratic formula.
We then substitute back to get our solutions: \( r = \frac{4 \pm 8 \sqrt{5}}{8} \), which simplifies to \( r = \frac{1}{2} \pm \sqrt{5} \).
Understanding how to simplify square roots helps to keep the final solutions in the simplest form, making them easier to understand and work with.