Question

Find the real solutions of each equation. $$ x^{6}-7 x^{3}-8=0 $$

Short answer

x = 2 and x = -1.

Step-by-step solution

- Introduce Substitution

Let’s make the substitution \(y = x^3\). This simplifies the equation to a quadratic form: y^2 - 7y - 8 = 0

- Solve the Quadratic Equation

We need to solve the quadratic equation y^2 - 7y - 8 = 0. This can be done using the quadratic formula:\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, a = 1, b = -7, and c = -8. Substitute these values into the formula:\[ y = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \]This gives us two solutions:\[ y_1 = \frac{7 + 9}{2} = 8 \]\[ y_2 = \frac{7 - 9}{2} = -1 \]

- Revert Substitution

Recall the substitution y = x^3. So:\[ x^3 = 8 \rightarrow x = \sqrt[3]{8} = 2 \]\[ x^3 = -1 \rightarrow x = \sqrt[3]{-1} = -1 \]

- List Real Solutions

The real solutions to the original equation x^6 - 7x^3 - 8 = 0 are x = 2 and x = -1.

Key concepts

Substitution Method

The substitution method in polynomial equations helps simplify complex equations. Here, the idea is to replace part of the equation with a different variable to make it easier to solve.

For example, in the given problem, we replace the variable expression with another variable:

• Original: \( x^6 - 7x^3 - 8 = 0 \)
• Substitution: Let \( y = x^3 \), transforming it to \( y^2 - 7y - 8 = 0 \)

By using substitution, we change a sixth-degree polynomial into a simpler quadratic equation. This helps us leverage simpler solving methods like the quadratic formula.

Quadratic Formula

Once we have a quadratic equation, we can use the quadratic formula to find the solutions. The quadratic formula is given by:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The variables represent the coefficients in the quadratic equation \( ax^2 + bx + c = 0 \). In the example problem:

• a = 1
• b = -7
• c = -8

Plugging these into the formula, we get:

\[ y = \frac{7 \pm \sqrt{49 + 32}}{2} = \frac{7 \pm \sqrt{81}}{2} = \frac{7 \pm 9}{2} \]
This offers two solutions:

• \( y_1 = 8 \)
• \( y_2 = -1 \)

Real Solutions

The real solutions of a polynomial equation are the values of the variable that satisfy the equation and are real numbers. In our example, after finding \( y_1 \) and \( y_2 \), we need to revert the substitution to find \( x \):

• For \( y_1 = 8 \): \( x^3 = 8 \) means \( x = \sqrt[3]{8} = 2 \)
• For \( y_2 = -1 \): \( x^3 = -1 \) means \( x = \sqrt[3]{-1} = -1 \)

Thus, the real solutions to the equation \( x^6 - 7x^3 - 8 = 0 \) are \( x = 2 \) and \( x = -1 \). These solutions are values of \( x \) that can be understood in real-life applications.

Polynomial Equations

Polynomial equations are equations involving a polynomial expression, a mathematical expression that involves sums of powers in one or more variables multiplied by coefficients.

For instance, the original problem is a sixth-degree polynomial equation given by:
\( x^6 - 7x^3 - 8 = 0 \)

Polynomial equations can have both real and complex solutions. Solving these equations can sometimes be simplified by transforming them into more manageable forms, such as using the substitution method.

• They can represent various real-life scenarios like physics problems, financial calculations, and engineering tasks.
• Understanding polynomials is crucial for advanced mathematical studies.