Problem 39 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No Real Solution. $$ x^{2}+x=4 $$

Step-by-Step Solution

Answer

The solutions are \( \frac{-1 + \sqrt{17}}{2} \) and \( \frac{-1 - \sqrt{17}}{2} \).

Step 1: Write the Equation in Standard Form

Start with the given equation: \( x^2 + x = 4 \). To use the quadratic formula, the equation must be in the form \( ax^2 + bx + c = 0 \). Rearrange the equation by subtracting 4 from both sides: \( x^2 + x - 4 = 0 \).

Step 2: Identify Coefficients

The standard form of the quadratic equation is \( ax^2 + bx + c = 0 \). For this equation, \( a = 1 \), \( b = 1 \), and \( c = -4 \).

Step 3: Use the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = 1 \), and \( c = -4 \) into the formula: \( x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-4)}}{2(1)} \).

Step 4: Calculate the Discriminant

The discriminant is the part of the quadratic formula under the square root: \( b^2 - 4ac \). Calculate it: \( 1^2 - 4(1)(-4) = 1 + 16 = 17 \). Since the discriminant is positive, there are two real solutions.

Step 5: Solve for the Roots

Substitute back the discriminant into the quadratic formula: \( x = \frac{-1 \pm \sqrt{17}}{2} \). This gives the two solutions: \( x = \frac{-1 + \sqrt{17}}{2} \) and \( x = \frac{-1 - \sqrt{17}}{2} \).

Key Concepts

Quadratic Equations

A quadratic equation is a special type of polynomial equation that can be expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The term with the highest power, \( x^2 \), gives it the name 'quadratic'. Solving these equations means finding the value(s) of \( x \) that make the equation true.
Quadratic equations can often appear in various real-world applications such as physics, engineering, and finance.
They can be solved by multiple methods, including:

• Factoring
• Completing the square
• Using the quadratic formula

In our example of \( x^2 + x = 4 \), we rearrange it to \( x^2 + x - 4 = 0 \) to fit the standard form. The quadratic formula enables us to find solutions effectively, especially when other methods are cumbersome.

Discriminant

The discriminant is a key component of the quadratic formula given by \( b^2 - 4ac \). It is the part under the square root in the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The value of the discriminant reveals a lot about the nature of the solutions to the quadratic equation.
The discriminant helps us determine:

• Positive Discriminant: Two distinct real solutions.
• Zero Discriminant: Exactly one real solution (a repeated root).
• Negative Discriminant: No real solutions (the solutions are complex).

In our specific case, the discriminant is \( 1^2 - 4(1)(-4) = 17 \). Since 17 is positive, the equation \( x^2 + x - 4 = 0 \) has two distinct real solutions.

Real Solutions

Real solutions are the values of \( x \) that satisfy the quadratic equation and fall within the real number system. In the context of the quadratic formula, real solutions occur when the discriminant (\( b^2 - 4ac \)) is greater than or equal to zero.
For the equation \( x^2 + x - 4 = 0 \), we calculated a positive discriminant of 17, resulting in two real solutions. These solutions are represented by:

• \( x = \frac{-1 + \sqrt{17}}{2} \)
• \( x = \frac{-1 - \sqrt{17}}{2} \)

These solutions are the points on the x-axis where the parabola described by the quadratic equation intersects, each indicating that they are real and distinct values from the set of real numbers. Real solutions are essential in practical applications as they represent real-world quantities.