Problem 49 - 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 complex: \(3 \pm i\).

Step 1: Simplify the Equation

First, observe that all the terms in the equation have a common factor of 2. We can simplify the equation by dividing each term by 2 to get: \[ x^2 - 6x + 10 = 0 \] This makes the equation easier to work with.

Step 2: Determine Method of Solution

The simplified quadratic equation is \(x^2 - 6x + 10 = 0\). Next, we decide whether to factor or use the quadratic formula. Since this expression does not factor easily into integer solutions, we will use the quadratic formula.

Step 3: Apply the Quadratic Formula

The quadratic formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -6 \), and \( c = 10 \) in our equation \(x^2 - 6x + 10 = 0\).

Step 4: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\) as follows: \[ (-6)^2 - 4 \cdot 1 \cdot 10 = 36 - 40 = -4 \] Since the discriminant is negative, the equation has no real solutions, only complex solutions.

Step 5: Solve for Complex Solutions

Using the quadratic formula we find the complex solutions: \[ x = \frac{-(-6) \pm \sqrt{-4}}{2 \cdot 1} = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} = 3 \pm i \] The solutions are \( 3 + i \) and \( 3 - i \).

Key Concepts

Factoring Quadratics

Factoring quadratics is one of the fundamental methods for solving quadratic equations. It involves expressing a quadratic equation in the form of

• \( ax^2 + bx + c = 0 \)

as a product of two binomials. This can make the equation much simpler to solve by finding values of \( x \) that make the expression equal zero. This method is very useful when the quadratic can be factored into real numbers.
However, not all quadratics factor easily. Here’s a process to factor a quadratic equation:

• Look for the greatest common factor (GCF) of the terms and factor it out, if possible.
• Transform the quadratic into a set of factors \( (x + p)(x + q) \) that reflects the given equation.
• Set each factor equal to zero and solve for \( x \).

In the original equation \( 2x^2 - 12x + 20 = 0 \), after dividing by 2, we are left with \( x^2 - 6x + 10 = 0 \). This expression does not easily factor into real numbers, which is why factoring was not used further.

Quadratic Formula

The quadratic formula is a universal method for solving any quadratic equation, even when factoring is impossible. This formula provides a straightforward way to solve for \(x\):

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The formula uses the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \), specifically \( a \), \( b \), and \( c \). Here are the steps:

• Identify the values of \( a \), \( b \), and \( c \) from your equation.
• Calculate the discriminant, \( b^2 - 4ac \).
• Use the discriminant to determine the nature of the roots. If it's positive, you'll have two real solutions. If zero, one real solution. If negative, two complex solutions.
• Plug these values into the formula to find your solutions for \( x \).

In our example, substituting \( a = 1 \), \( b = -6 \), and \( c = 10 \) into the formula allows us to find solutions even though the equation cannot be factored into real numbers.

Complex Solutions

Complex solutions occur when the discriminant \( b^2 - 4ac \) is negative. A negative discriminant means that the square root within the quadratic formula yields an imaginary number. This is where

• \( i \)

comes into play, where \( i \) is defined as

• \( \sqrt{-1} \)

To compute complex solutions:

• Use the quadratic formula as usual.
• When encountering a negative discriminant, convert the square root of the negative number into \( i \), leading to imaginary components in the solution.
• The solutions will have the form
  • \( a \pm bi \)
  , where \( a \) is the real part and \( bi \) is the imaginary part of the complex solution.

For the equation \( x^2 - 6x + 10 = 0 \), the discriminant is \(-4\), resulting in imaginary numbers for solutions:

• \( 3 + i \)
• \( 3 - i \)

. These indicate that the solutions to the quadratic equation include imaginary numbers, leading to complex numbers as results.