Chapter 1: Problem 68
Solve each cubic equation using factoring and the quadratic formula. $$x^{3}-27=0$$
Short Answer
Expert verified
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 by step solution
01
Identify the cubic equation
The given equation is \(x^{3}-27=0\). Recognize it as a difference of cubes.
02
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 \(b = 3\).
03
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.
04
Solve the linear factor
Set the linear factor equal to zero: \(x - 3 = 0\). Solving this gives \(x = 3\).
05
Solve the quadratic factor
Set the quadratic factor equal to zero to solve for the remaining solutions: \(x^{2} + 3x + 9 = 0\).
06
Apply the quadratic formula
Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\) with \(a = 1\), \(b = 3\), and \(c = 9\).
07
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.
08
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 \(x = \frac{-3 - 3i\sqrt{3}}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
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 \((x - 3)(x^{2} + 3x + 9) = 0\).
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: \(x^{2} + 3x + 9 = 0\).
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.
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:
Hence, our complex roots are: \(x = \frac{-3 + 3i\sqrt{3}}{2}\) and \(x = \frac{-3 - 3i\sqrt{3}}{2}\).
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 \(x = \frac{-3 - 3i\sqrt{3}}{2}\).