Problem 74 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the equation using any convenient method. $$x^{2}+4 x=-\frac{19}{4}$$

Step-by-Step Solution

Answer

The equation has no real solutions because the discriminant ( \( b^{2} - 4ac \)) is negative.

Step 1: Rewrite the Equation

First, the equation should be brought into the standard form of a quadratic equation. This is done by adding \( \frac{19}{4} \) to both sides to get the equation \( x^{2}+4 x+\frac{19}{4} = 0 \).

Step 2: Identify a, b, and c

From the standard form of a quadratic equation \( ax^2 + bx + c = 0 \), identify \( a, b, c \) as \( 1, 4, \frac{19}{4} \) respectively.

Step 3: Use Quadratic Formula

The quadratic formula is \( x=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \). Substituting \( a, b, c \) from step 2 we can find the possible values of \( x \) using this formula. Calculating the root under the square root sign \( b^{2}-4ac \) = \( 4^2 - 4*1*\frac{19}{4} \) gives \( -3 \), but root of negative number doesn't exist in real number system, so there is no real solution to this equation.

Key Concepts

Quadratic Formula

The quadratic formula, a pivotal tool in algebra, is used to determine the roots of a quadratic equation. This formula is expressed as \( x=\frac{{-b \pm \sqrt{{b^{2}-4ac}}}}{{2a}} \). It provides a systematic way to find the solutions by simply plugging in the values of \(a\), \(b\), and \(c\), which are coefficients and constant term from the standard form \( ax^2 + bx + c = 0 \).

To apply it, one must perform several steps: firstly, ensure that the equation is in the standard form; then identify the values of \(a\), \(b\), and \(c\); and finally, verify the discriminant, \(b^{2}-4ac\), which can reveal if there are real solutions or not. If the discriminant is positive, you'll find two real and distinct solutions; if it's zero, there's exactly one real solution; and if it's negative, as in the textbook example, there will be no real solutions, which we'll discuss further in the next section.

No Real Solution

Sometimes, when applying the quadratic formula, you may encounter situations where there is no real solution. This occurs if the discriminant \(b^{2}-4ac\) is negative. Since the square root of a negative number is not a real number, the quadratic equation in such cases does not intersect the x-axis and, thus, has no real x-intercepts.

In the context of the given problem, we calculated the discriminant to be \(4^2 - 4 \times 1 \times \frac{19}{4} = -3\), which is less than zero. Therefore, we conclude there are no real solutions. However, in the complex number system, non-real solutions exist and are represented as a combination of a real part and an imaginary unit \(i\), which is defined as \(\sqrt{-1}\). Thus, the solutions are complex numbers, not real numbers.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is one of the most fundamental concepts when dealing with quadratics. It is expressed as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). Equations must be rearranged into this form to solve them using most methods, including the quadratic formula.

The original problem presented the equation \(x^{2}+4x=-\frac{19}{4}\). To solve it using the quadratic formula or by factoring, the first step is to rewrite it into standard form, resulting in \(x^{2}+4x+\frac{19}{4}=0\). This maneuver allows us to identify \(a=1\), \(b=4\), and \(c=\frac{19}{4}\), and to proceed effectively to solve the equation. Remember, without the correct standard form, it's challenging to analyze or solve quadratic equations accurately.