Problem 100 - Algebra and Trigonometry

Question

Solve equation by the method of your choice. $$ x^{2}-4 x+29=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2}-4x+29=0\) are \(x = 2 + 10i\sqrt{6}\) and \(x = 2 - 10i\sqrt{6}\).

Step 1: Identify the coefficients

First recognize the equation is in the form \(ax^{2}+bx+c=0\). In this case, \(a=1\), \(b=-4\), and \(c=29\).

Step 2: Substitute coefficients in the quadratic formula

The quadratic formula is \(-b±\sqrt{b^{2}-4ac}/2a\). Substitute the identified values into it: \(x=4±\sqrt{(-4)^{2}-4*29}/2*1\)

Step 3: Simplify the expression under the square root

Inside the square root, you have \(16 - 4*29\), which simplifies to -96.

Step 4: Simplify the equation

Since the term inside the square root is negative, the solutions will be complex. The square root of -96 can be written as \(10i\sqrt{96}\). So, the solutions are \(x = 4± 10i\sqrt{96}\).

Step 5: Find two solutions

Finally, the quadratic equation has two solutions: \(x = 2 + 10i\sqrt{6}\) and \(x = 2 - 10i\sqrt{6}\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerhouse of algebra, providing a reliable method to find the roots of any quadratic equation. Quadratic equations typically look like this:
\(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a \eq 0\). The quadratic formula itself is derived from completing the square in the equation and is as follows:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
By inserting the coefficients of your equation into this formula, you can solve for \(x\) to find the roots of the polynomial, which can be either real or complex numbers. This formula is celebrated for its ability to handle even when the discriminant, here \(b^2 - 4ac\), is negative, leading to complex solutions.

Complex Numbers

Complex numbers expand the horizon beyond the real number system. They consist of a real part and an imaginary part and are written in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, satisfying the equation \(i^2 = -1\).

When solving quadratic equations, if the discriminant (the part under the square root in the quadratic formula) is negative, we enter the realm of complex numbers. For instance, the square root of a negative number, such as \(\sqrt{-96}\), cannot be represented on the real number line. Instead, it is expressed in terms of \(i\), like \(\sqrt{-96} = 10i\sqrt{6}\). It is essential to understand that complex numbers are not 'less real' than real numbers; they're just different and are crucial for solving certain types of mathematical problems.

Roots of Polynomial

The roots of a polynomial are the values of \(x\) that make the polynomial equal to zero. In the context of a quadratic equation, which is a second-degree polynomial, there are typically two roots. These roots can be real numbers or complex numbers, depending on the discriminant of the quadratic equation.

When the discriminant is positive, we get two different real roots. If it's zero, we get exactly one real root (also called a repeated or double root). However, if the discriminant is negative, which signifies there are no real roots, the quadratic formula gives us two complex roots, as in our example equation \(x^2 - 4x + 29 = 0\), where the roots are \(x = 2 + 10i\sqrt{6}\) and \(x = 2 - 10i\sqrt{6}\). These roots are the solutions to our polynomial and are vital to fully understand its behavior.