Problem 68 - Precalculus Student Solutions Manual 5th

Question

Solve each cubic equation using factoring and the quadratic formula. $$x^{3}-27=0$$

Step-by-Step Solution

Answer

The solutions to \(x^{3} - 27 = 0\) are \(x = 3\), \(x = \frac{-3 + 3i\sqrt{3}}{2}\), and \(x = \frac{-3 - 3i\sqrt{3}}{2}\).

Step 1: Identify the cubic equation

The given equation is \(x^{3}-27=0\). Recognize it as a difference of cubes.

Step 2: Rewrite as a difference of cubes

The equation can be written as \(x^{3} - 3^{3} = 0\). This fits the form \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\). Here, \(a = x\) and \(b = 3\).

Step 3: Factor the difference of cubes

Applying the difference of cubes formula, we get: \((x - 3)(x^{2} + 3x + 9) = 0\). So, we factored the cubic equation into a linear term and a quadratic term.

Step 4: Solve the linear factor

Set the linear factor equal to zero: \(x - 3 = 0\). Solving this gives \(x = 3\).

Step 5: Solve the quadratic factor

Set the quadratic factor equal to zero to solve for the remaining solutions: \(x^{2} + 3x + 9 = 0\).

Step 6: Apply the quadratic formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b = 3\), and \(c = 9\).

Step 7: Calculate the discriminant

The discriminant is \(b^{2} - 4ac = 3^{2} - 4(1)(9) = 9 - 36 = -27\). Since the discriminant is negative, there are no real roots, only complex roots.

Step 8: Solve for complex roots

Plug the values into the quadratic formula: \[x = \frac{-3 \pm \sqrt{-27}}{2} = \frac{-3 \pm 3i\sqrt{3}}{2} \]. This simplifies to \(x = \frac{-3 + 3i\sqrt{3}}{2}\) and \(x = \frac{-3 - 3i\sqrt{3}}{2}\).

Key Concepts

Difference of Cubes

In mathematics, the 'difference of cubes' is a technique used to simplify and solve problems involving cubic equations like the one given in our exercise: \(x^{3}-27=0\). Recognizing a cubic equation can make solving it easier.

To identify a difference of cubes, you should look for expressions in the form \(a^{3} - b^{3}\). The formula to factor this is: \(a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2})\).

Let's apply this to our equation:

• Here, \(a = x\) and \(b = 3\) because \(3^{3}=27\).
• Using the formula, we rewrite \(x^{3} - 3^{3} = 0\) as \((x - 3)(x^{2} + 3x + 9) = 0\).

Through this factorization, we've broken down the cubic equation into a simpler linear term \(x-3\) and a quadratic term \(x^{2}+3x+9\). This makes it much easier to solve.

Quadratic Formula

The quadratic formula is a powerful tool to solve quadratic equations, especially when they can't be easily factored. A standard quadratic equation is in the form \( ax^2 + bx + c = 0 \). The quadratic formula is: \[x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a}\].

Let's use the quadratic formula on our quadratic term from the previous section: \(x^{2} + 3x + 9 = 0\).

• Here, \(a = 1\), \(b = 3\), and \(c = 9\).
• Calculating the discriminant \( b^2 - 4ac = 3^2 - 4(1)(9) = 9 - 36 = -27 \).
• A negative discriminant means that the solutions will be complex roots, which leads us to our next concept.

Simplifying the quadratic formula, we get complex roots.

Complex Roots

When we solve quadratic equations with a negative discriminant, we get complex roots. Complex roots include an imaginary part, indicated by \(i\), where \(i = \sqrt{-1}\).

From the earlier section, we calculated the discriminant \( 3^2 - 4(1)(9) = -27 \). Finding the roots involves:

• Plugging the values into the quadratic formula: \(x = \frac{-3 \pm \sqrt{-27}}{2}\).
• Simplifying under the square root: \(\sqrt{-27} = 3i\sqrt{3}\).
• So, the solutions are: \(x = \frac{-3 \pm 3i\sqrt{3}}{2}\).

Hence, our complex roots are: \(x = \frac{-3 + 3i\sqrt{3}}{2}\) and \(x = \frac{-3 - 3i\sqrt{3}}{2}\).