Problem 63 - Algebra and Trigonometry

Question

\(61-70\) . Find all solutions, real and complex, of the equation. $$ x^{3}+x^{2}+x=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \), \( x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \), \( x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \).

Step 1: Factor out the greatest common factor

The equation is given as \( x^3 + x^2 + x = 0 \). Notice that each term of the polynomial has an \( x \) in it. Factor out the common factor \( x \) from each term: \[ x(x^2 + x + 1) = 0 \]

Step 2: Solve for the first real root

From the factorization in Step 1, we separate the equation into simpler parts. The first equation is \( x = 0 \), which is a straightforward real root.

Step 3: Solve the quadratic equation

Now, look at the quadratic equation \( x^2 + x + 1 = 0 \). To find the roots, apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In our equation, \( a = 1 \), \( b = 1 \), \( c = 1 \).

Step 4: Calculate the discriminant

Compute the discriminant \( \Delta = b^2 - 4ac \): \[ \Delta = 1^2 - 4 \times 1 \times 1 = 1 - 4 = -3 \]Since the discriminant is negative, the roots will be complex.

Step 5: Compute the complex roots

Using the quadratic formula with the computed discriminant, the roots are: \[ x = \frac{-1 \pm \sqrt{-3}}{2 \times 1} \]These can be expressed as: \[ x = \frac{-1 \pm i \sqrt{3}}{2} \]This results in two complex roots, \( x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \) and \( x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \).

Key Concepts

Complex Roots in a Polynomial Equation

To fully understand complex roots, let's delve into the scenario when our polynomial equation doesn't yield simple real number solutions. A polynomial can sometimes have roots that are not just real numbers, but complex numbers as well. This happens when the part under the square root, known as the discriminant, in the quadratic formula is negative.

In the given exercise, the equation is factored into a linear polynomial and a quadratic polynomial:

• Linear: \( x = 0 \)
• Quadratic: \( x^2 + x + 1 = 0 \)

For the quadratic equation, computing the discriminant (\( \Delta \)) shows it is negative. Thus, complex roots arise. These roots always occur in conjugate pairs, meaning if one root is \( a + bi \), the other will be \( a - bi \). Remember, the 'i' stands for the imaginary unit, where \( i^2 = -1 \). This is critical for representing complex numbers—a combination of real and imaginary parts.

The solutions \( x = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \) and \( x = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \) are our complex conjugates for this particular quadratic equation.

Using the Quadratic Formula

The quadratic formula is a fundamental tool in algebra for finding the roots of a quadratic equation. Any equation of the form \( ax^2 + bx + c = 0 \) can be solved using the formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

In our solution, the quadratic part of the equation was \( x^2 + x + 1 \), where \( a = 1 \), \( b = 1 \), and \( c = 1 \). By plugging these values into the quadratic formula, we systematically calculate the roots of this quadratic equation. It's like a cookie-cutter process for finding solutions, where each term fits into place based on the equation's coefficients.

A key detail when using the quadratic formula is the discriminant, \( b^2 - 4ac \). This portion decides the nature of the roots:

• Positive discriminant: two distinct real roots
• Zero discriminant: one repeated real root
• Negative discriminant: two distinct complex roots

In our exercise, since the discriminant was \( -3 \), it confirmed the presence of complex roots.

Understanding Real Roots

Real roots are solutions that are actual numbers on the number line, as opposed to complex numbers which involve the imaginary unit \( i \). In the given exercise, we identify a real root from the equation \( x^3 + x^2 + x = 0 \) by factoring out the common factor \( x \), resulting in \( x(x^2 + x + 1) = 0 \). This gives us a real, tangible solution: \( x = 0 \).

Real roots are straightforward—essentially where the graph of the equation touches or intersects the x-axis. They provide a direct answer without involving complex numbers or imaginary units.

Recognizing and factoring to find these roots is crucial for solving polynomial equations efficiently. It allows us to simplify the equation before considering more complex methods like the quadratic formula. Once the polynomial is partially and easily reduced, solving for real roots becomes a simpler task of setting each factor equal to zero.