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Magazine Chapter 1: Problem 70
Solve each cubic equation using factoring and the quadratic formula. $$x^{3}+64=0$$
Short Answer
Expert verified
The solutions are \(x = -4\), \(x = 2 + 2i\sqrt{3}\), and \(x = 2 - 2i\sqrt{3}\).
Step by step solution
01
Identify the Cube Terms
The given equation is \(x^{3} + 64 = 0\). Notice that 64 can be written as \(4^3\). Therefore, the equation is of the form \(x^3 + a^3 = 0\) where \(a = 4\).
02
Apply Sum of Cubes Formula
The equation \(x^3 + 64 = 0\) can be factored using the sum of cubes formula: \(x^3 + a^3 = (x + a)(x^2 - ax + a^2)\). Substituting \(a = 4\), we get: \(x^3 + 4^3 = (x + 4)(x^2 - 4x + 16)\).
03
Set Each Factor to Zero
Set each factor in the product \((x + 4)(x^2 - 4x + 16)\) equal to zero. Solve for each factor: \(x + 4 = 0\) and \(x^2 - 4x + 16 = 0\).
04
Solve the Linear Equation
Solving \(x + 4 = 0\), we get \(x = -4\).
05
Use the Quadratic Formula
To solve \(x^2 - 4x + 16 = 0\) use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -4\), and \(c = 16\).
06
Calculate the Discriminant
Calculate the discriminant: \(b^2 - 4ac = (-4)^2 - 4(1)(16) = 16 - 64 = -48\). Since the discriminant is negative, the quadratic equation has two complex solutions.
07
Find Complex Solutions
Using the quadratic formula with the negative discriminant: \(x = \frac{4 \pm \sqrt{-48}}{2}\). Simplify the square root part: \(\sqrt{-48} = \sqrt{-1 \cdot 16 \cdot 3} = 4i\sqrt{3}\). So, \(x = \frac{4 \pm 4i\sqrt{3}}{2} = 2 \pm 2i\sqrt{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
factoring
Factoring is a key skill in solving polynomial equations, especially cubic ones. Factoring involves breaking down a complex expression into simpler terms, or 'factors,' that multiply together to give the original expression. In the context of the exercise, the equation is given as \[ x^3 + 64 = 0 \], and it needs to be factored to solve for x. Notice that 64 is a perfect cube \( 4^3 \). Therefore, the equation fits the sum of cubes formula, which can help us factor it into simpler binomial and trinomial expressions. Here, the factored form is: \[ (x + 4)(x^2 - 4x + 16) = 0 \]. Once factored, setting each factor to zero allows us to solve for x one by one.
sum of cubes formula
The sum of cubes formula is a special factoring formula used to factor expressions that are the sum of two perfect cubes. The formula is: \[ x^3 + a^3 = (x + a)(x^2 - ax + a^2) \]. In our exercise, the equation \( x^3 + 64 = 0 \) fits this format because \( 64 \) is \( 4^3 \). Using the sum of cubes formula, the factorization becomes: \[ x^3 + 64 = (x + 4)(x^2 - 4x + 16) \]. This simplifies the cubic equation into a product of a linear factor and a quadratic factor, making it easier to solve for x by setting each factor equal to zero.
quadratic formula
After factoring the cubic equation using the sum of cubes formula, we end up with a quadratic equation: \( x^2 - 4x + 16 = 0 \). To solve this quadratic equation, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. In our exercise, \( a = 1 \), \( b = -4 \), and \( c = 16 \). Plugging these values into the formula, we get:\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(16)}}{2(1)} = \frac{4 \pm \sqrt{16 - 64}}{2} = \frac{4 \pm \sqrt{-48}}{2} \]. This gives us the solutions for the quadratic equation, involving complex numbers due to the negative discriminant.
complex solutions
Complex solutions arise when the discriminant (\( b^2 - 4ac \)) of a quadratic equation is negative. In our exercise, we found \( b^2 - 4ac = -48 \), which is a negative number. This means the quadratic equation has two complex solutions. To find these solutions, we need to handle the square root of a negative number using the imaginary unit \( i \), where \( i = \sqrt{-1} \). The quadratic formula simplifies to:\[ x = \frac{4 \pm \sqrt{-48}}{2} = \frac{4 \pm 4i\sqrt{3}}{2} = 2 \pm 2i\sqrt{3} \]. These are the complex solutions to our quadratic equation. Complex solutions usually come in conjugate pairs, making them essential when dealing with polynomials and quadratic equations.
