Problem 58 - Algebra and Trigonometry

Question

The given equation involves a power of the variable. Find all real solutions of the equation. \(6 x^{2}+100=0\)

Step-by-Step Solution

Answer

The equation \(6x^2 + 100 = 0\) has no real solutions.

Step 1: Understand the Equation

The given equation is \(6x^2 + 100 = 0\). It is a quadratic equation in the form \(ax^2 + c = 0\), where \(a = 6\) and \(c = 100\). Our goal is to find the values of \(x\) that satisfy this equation.

Step 2: Isolate the Quadratic Term

To make the equation easier to solve, isolate the term with the power of the variable. Subtract 100 from both sides of the equation to get \(6x^2 = -100\).

Step 3: Solve for the Variable

Next, we need to solve for \(x^2\) by dividing both sides by 6. This simplifies our equation to \(x^2 = -\frac{100}{6}\), which can be further simplified to \(x^2 = -\frac{50}{3}\).

Step 4: Consider Real Solutions

Considering that \(x^2 = -\frac{50}{3}\), we notice that the square of a real number cannot be negative. Since there are no real numbers that satisfy this equation, there are no real solutions for \(x\).

Key Concepts

Real Solutions

In the world of mathematics, particularly when dealing with quadratic equations, finding real solutions is often the primary goal. A real solution is a number that does not involve any imaginary components.
In simple terms, it's a number that you can find on the regular number line you're familiar with. For a quadratic equation such as the one given in the exercise, which takes the form of \( ax^2 + c = 0 \), we aim to find real numbers that satisfy this equality.
However, in this specific case, the equation resolves to \( x^2 = -\frac{50}{3} \). Since the square of any real number is always non-negative, a negative result like \(-\frac{50}{3}\) indicates that real solutions do not exist. Thus, this equation does not have any real solutions.

• Real solutions are numbers without imaginary parts.
• They are values found on the standard number line.
• Negative square values mean no real solutions exist.

Solving Quadratic Equations

Quadratic equations are fundamental in algebra and have the general form \( ax^2 + bx + c = 0 \). Our exercise simplifies to a special case where \( b = 0 \). To solve \( 6x^2 + 100 = 0 \), we start by isolating the quadratic term, \( x^2 \).
This is done by subtracting 100 from both sides to obtain \( 6x^2 = -100 \). Dividing both sides by 6 gives \( x^2 = -\frac{50}{3} \). The point is to solve for \( x \) by taking the square root of both sides. Yet, here lies the catch:
The expression \( x^2 = -\frac{50}{3} \) suggests taking the square root of a negative number, which leads to a different type of solution. This type of scenario introduces the concept of imaginary numbers.

• Quadratic equations typically have two solutions.
• Isolating \( x^2 \) is crucial in finding these solutions.
• Negative results under the square introduce complex numbers.

Imaginary Numbers

When solving quadratic equations, especially those yielding negative values for \( x^2 \), we encounter imaginary numbers. Imaginary numbers are rooted in the concept that the square root of a negative number cannot be a real number.
To represent such values, mathematicians introduced \(i\), defined as \( i = \sqrt{-1} \). So in a case like \( x^2 = -\frac{50}{3} \), the solution involves taking the square root of both sides, utilizing \( i \). Consequently, the solutions are better expressed in terms of imaginary numbers like \( x = \sqrt{\frac{50}{3}} i \) and \( x = -\sqrt{\frac{50}{3}} i \).

• Imaginary numbers use \(i\) to deal with negative square roots.
• They extend real numbers to "complex numbers."
• The form of complex numbers is often \( a + bi \), where \( a \) is a real number and \( b \) is a coefficient of \( i \).

This extension allows solving equations that have no real solutions, enriching our understanding and capability in mathematical problem-solving.