Problem 39 - Algebra 1: Concepts and Skills

Question

Solve the equation or write no real solution. Write the solutions as integers if possible. Otherwise, write them as radical expressions. $$ x^{2}+36=0 $$

Step-by-Step Solution

Answer

The given equation does not have any real solutions

Step 1: Identify the quadratic equation

The given equation is \(x^{2}+36=0\). Here, the coefficients are a=1 (coefficient of \(x^{2}\)), b=0 (coefficient of x) and c=36 (constant term).

Step 2: Evaluate the determinant

The determinant of a quadratic equation is found using the formula \(b^{2}-4ac\). Substituting the values, we get \(0^{2}-4*1*36\), which simplifies to -144.

Step 3: Check the determinant for solutions

Since the determinant is negative, this suggests that the quadratic equation does not have any real solutions. The solutions, if any, will be imaginary or complex numbers.

Key Concepts

Determinant

The determinant is a key concept in understanding quadratic equations. It helps us determine the nature of the solutions we can expect from the equation. The determinant of a quadratic equation is calculated using the formula:

• \(b^2 - 4ac\) - where \(a\), \(b\), and \(c\) are the coefficients of the terms in the equation \(ax^2 + bx + c = 0\).

For the example on hand: \(x^2 + 36 = 0\), the calculated determinant is:

• \(b^2 - 4ac = 0^2 - 4 \times 1 \times 36 = -144\)

This calculation shows that the determinant is negative. A negative determinant indicates that the solutions to the quadratic equation are not real numbers. Understanding this helps us anticipate whether solutions can be directly calculated or if they require handling complex numbers, as is the case here.

No Real Solutions

When the determinant of a quadratic equation is negative, it means there are no real solutions. Real solutions are numbers we can find on the number line. A negative determinant signifies that the curve of the quadratic equation does not intersect the x-axis, making real solutions impossible.
In the given equation \(x^2 + 36 = 0\), the determinant \(-144\) confirms this scenario. The graph of \(x^2 = -36\) is above the x-axis, thus no real x-intercepts exist. Instead, the solutions involve complex or imaginary numbers.

• If the determinant is positive, there would be two distinct real solutions.
• If the determinant is zero, there would be exactly one real solution, known as a repeated or double root.

Complex Numbers

Complex numbers become crucial when dealing with quadratic equations having a negative determinant. A complex number is an expression involving both a real part and an imaginary part, written as \(a + bi\).

• Here, \(a\) is the real part and \(b\) is the imaginary part \((i)\).

For the equation \(x^2 + 36 = 0\), we solve for \(x\) using:

• \(x = \pm\sqrt{-36}\).

This results in an imaginary number because you cannot take the square root of a negative number using real numbers. Instead, we write it as:

• \(x = \pm6i\) - because \(\sqrt{-1} = i\).

Understanding that \(i\) is the fundamental unit of imaginary numbers helps us solve such problems. This way, when a real solution isn't possible, complex numbers provide a comprehensive answer.