Problem 61 - Algebra and Trigonometry

Question

55–64 ? Find all solutions, real and complex, of the equation. $$ x^{6}-9 x^{3}+8=0 $$

Step-by-Step Solution

Answer

The solutions are \( \{ 1, 2, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}}, 2e^{i\frac{2\pi}{3}}, 2e^{-i\frac{2\pi}{3}} \} \).

Step 1: Substitute variables

Let us make the substitution \( y = x^3 \). This transforms the given equation to a quadratic equation in terms of \( y \): \( y^2 - 9y + 8 = 0 \).

Step 2: Solve the quadratic equation

We use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In this equation, \( a = 1 \), \( b = -9 \), and \( c = 8 \). So,\[y = \frac{9 \pm \sqrt{81 - 32}}{2} = \frac{9 \pm \sqrt{49}}{2} = \frac{9 \pm 7}{2}.\]This gives us two solutions: \( y = 8 \) and \( y = 1 \).

Step 3: Solve for \( x \) when \( y = 8 \)

Since \( y = x^3 = 8 \), taking the cube root of both sides gives us \( x = \sqrt[3]{8} = 2 \).

Step 4: Solve for \( x \) when \( y = 1 \)

Since \( y = x^3 = 1 \), taking the cube root of both sides gives us \( x = \sqrt[3]{1} = 1 \).

Step 5: Consider complex solutions

Because \( x^3 = 8 \) and \( x^3 = 1 \) are cubic equations, they have three roots each (one real and two complex for non-zero values). For \( x^3 = 8 \), the cube roots are \( 2 \), \( 2e^{i\frac{2\pi}{3}} \), and \( 2e^{-i\frac{2\pi}{3}} \). For \( x^3 = 1 \), the cube roots are \( 1 \), \( e^{i\frac{2\pi}{3}} \), and \( e^{-i\frac{2\pi}{3}} \).

Step 6: Compile all solutions

Combine the real and complex roots found:- From \( x^3 = 8 \): \( 2, 2e^{i\frac{2\pi}{3}}, 2e^{-i\frac{2\pi}{3}} \)- From \( x^3 = 1 \): \( 1, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}} \)Thus, the complete set of solutions is \( \{ 1, 2, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}}, 2e^{i\frac{2\pi}{3}}, 2e^{-i\frac{2\pi}{3}} \} \).

Key Concepts

Complex Numbers

Understanding complex numbers is key to solving many polynomial equations. Complex numbers are numbers that have a real part and an imaginary part. The imaginary unit is represented by \(i\), where \(i^2 = -1\). So a general complex number is written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part.

• The role of complex numbers in equations is crucial when the polynomial has roots that are not real. This happens when the solutions require taking the square root of a negative number.
• In our example, the complex roots arise when finding the cube roots of numbers like \(8\) and \(1\) because these equations involve \(e^{i\frac{2\pi}{3}}\) and \(e^{-i\frac{2\pi}{3}}\), which are complex numbers.

To find all the roots of a cubic polynomial like the equations \(x^3 = 8\) or \(x^3 = 1\), you use both real and complex solutions to ensure all possibilities are covered.

Roots of Equations

Finding the roots of equations is a fundamental aspect of algebra. Roots of an equation are the values of the variable that satisfy the equation. In polynomial equations, these are the values where the polynomial equals zero.

• For any polynomial, the degree of the polynomial tells you the number of roots. For example, a cubic polynomial will generally have three roots, which can be real or complex.
• In our exercise, once the equation was transformed into a quadratic form using substitution, it was found that the roots of \(y^2 - 9y + 8 = 0\) are \(y = 8\) and \(y = 1\).

After finding \(y\), reverting back to \(x\) gives the original roots. For \(x^3 = 8\), the roots are \(2, 2e^{i\frac{2\pi}{3}}, 2e^{-i\frac{2\pi}{3}}\). For \(x^3 = 1\), the roots are \(1, e^{i\frac{2\pi}{3}}, e^{-i\frac{2\pi}{3}}\). Combining these gives all possible solutions for the initial equation.

Quadratic Formula

The quadratic formula is a method used to find the solutions of a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula directly gives the roots of the quadratic equation after substituting the values of coefficients \(a\), \(b\), and \(c\).

• The expression under the square root, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots: real and distinct, real and equal, or complex.
• In our exercise, the discriminant is \(81 - 32 = 49\), a perfect square, indicating that all roots are real.

Using the quadratic formula not only streamlined solving \(y^2 - 9y + 8 = 0\) into finding \(y = 8\) and \(y = 1\), but also demonstrated a thorough understanding of root-finding methods in polynomials.