Problem 49 - Brief Applied Calculus

Question

Solve each equation by factoring or the Quadratic Formula, as appropriate. $$ 2 x^{2}-12 x+20=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = 5\).

Step 1: Identify the Type of Equation

The equation given is a quadratic equation in the form \( ax^2 + bx + c = 0 \). In this case, \( a = 2 \), \( b = -12 \), and \( c = 20 \).

Step 2: Check for Factoring

To see if the quadratic equation can be factored easily, look for two numbers that multiply to \( a \cdot c = 2 \cdot 20 = 40 \) and add up to \( b = -12 \). The numbers \(-10\) and \(-2\) multiply to \(40\) and add to \(-12\).

Step 3: Rewrite the Quadratic Equation

Rewrite the equation using the numbers identified in step 2: \[ 2x^2 - 10x - 2x + 20 = 0 \] This splits the middle term into two terms whose coefficients are the numbers found.

Step 4: Factor by Grouping

Group the terms to factor by grouping: \[ 2x(x - 5) - 2(x - 5) = 0 \]. Notice both groups have a common factor of \((x - 5)\).

Step 5: Extract Common Factors

Factor out the common factor: \[ (2x - 2)(x - 5) = 0 \]. Additionally, factor out a 2 from \(2x - 2\), resulting in: \[ 2(x - 1)(x - 5) = 0 \].

Step 6: Solve for the Variable

Set each factor equal to zero and solve: \(2(x - 1) = 0 \) implies \(x = 1\), \((x - 5) = 0 \) implies \(x = 5\).

Step 7: Verify the Solutions

Substitute each solution back into the original equation to verify correctness. Both \(x = 1\) and \(x = 5\) satisfy the equation.

Key Concepts

Factoring Quadratics

Factoring quadratics is a method used to solve quadratic equations by expressing them as a product of simpler expressions. Not all quadratic equations can be solved by factoring, but when they can, it often provides a quick and clear solution.

• First, identify the form of the quadratic equation, which is usually written as \( ax^2 + bx + c = 0 \).
• Next, find two numbers that multiply to \( a \cdot c \) and add to \( b \). For example, in the equation \( 2x^2 - 12x + 20 = 0 \), we identify \( a = 2 \), \( b = -12 \), and \( c = 20 \). Here, \( a \cdot c \) equals 40.
• The numbers \(-10\) and \(-2\) are identified because they multiply to 40 and add to -12.
• Rewrite the middle term using these numbers, then group the equation and factor by grouping.

Factoring by grouping involves arranging terms into groups and then factoring out the common factor from each group. This technique helps to simplify the equation into a product of binomials. In our example, the factorization leads to \( 2(x-1)(x-5) = 0 \), which makes it straightforward to identify the solutions.

Quadratic Formula

The quadratic formula is a universal method applicable to all quadratic equations, especially when factoring is challenging. The standard formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here is how you can apply it:

• Identify values of \( a \), \( b \), and \( c \) from the standard quadratic form \( ax^2 + bx + c = 0 \).
• Substitute these values into the quadratic formula. The expression under the square root, \( b^2 - 4ac \), is called the discriminant.

The discriminant is crucial as it indicates the nature of the roots:

• If positive, the quadratic has two distinct real roots.
• If zero, there is one real root (a repeated root).
• If negative, the roots are complex and not real.

For example, using the quadratic formula may not have been necessary for \( 2x^2 - 12x + 20 = 0 \), as it factors easily. However, the formula guarantees finding the roots efficiently even when factoring is not apparent.

Roots of Equations

The roots of an equation are the solutions that make the equation true. For a quadratic equation, these roots can be found using methods such as factoring or the quadratic formula. Their significance is as follows:

• Roots represent the values where the quadratic function intersects the x-axis.
• Finding roots means determining values of \( x \) that make the quadratic expression zero.

In the provided example, solving \( 2(x-1)(x-5) = 0 \) reveals the roots \( x = 1 \) and \( x = 5 \). Here's how each solution is achieved:

• Setting \( (x-1) = 0 \) gives \( x = 1 \).
• Setting \( (x-5) = 0 \) gives \( x = 5 \).

It's essential to substantiate these roots by substituting them back into the original equation to ensure they balance the equation to zero. Verifying solutions helps confirm that the roots are correct, creating confidence in problem-solving abilities. Understanding these roots is crucial as they reveal important properties about the function, such as its intercepts, and assist in graphing and analyzing quadratic relationships.