Chapter 6: Problem 37
Solve each equation. See Examples 7 through \(9 .\) $$ x^{3}-12 x^{2}+32 x=0 $$
Short Answer
Expert verified
The solutions are \( x = 0 \), \( x = 4 \), and \( x = 8 \).
Step by step solution
01
Factor Out the Common Term
The given equation is \( x^3 - 12x^2 + 32x = 0 \). Notice all terms include \( x \), so the first step is to factor the common \( x \) out: \( x(x^2 - 12x + 32) = 0 \). Now we have one factor as \( x \) and another quadratic equation.
02
Solve for x from the First Factor
From the factor \( x \), we set it equal to zero: \( x = 0 \). This is our first solution of the equation.
03
Solve the Quadratic Equation
Now solve the quadratic equation \( x^2 - 12x + 32 = 0 \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -12 \), and \( c = 32 \).
04
Calculate the Discriminant
First, find the discriminant: \( (-12)^2 - 4 \times 1 \times 32 = 144 - 128 = 16 \). This is positive, indicating two real solutions.
05
Apply the Quadratic Formula
Now use the quadratic formula: \( x = \frac{12 \pm \sqrt{16}}{2} \). Simplifying gives \( x = \frac{12 \pm 4}{2} \).
06
Find the Two Real Solutions
Calculate the two solutions: \( x = \frac{12 + 4}{2} = 8 \) and \( x = \frac{12 - 4}{2} = 4 \). These are the solutions to the quadratic equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring
Factoring is a powerful tool in solving equations, especially when dealing with polynomials. When you factor, you're essentially breaking down a complex expression into simpler components, called factors, that when multiplied together give back the original expression. In the example equation, we first observed that each term contained the variable \( x \), which allowed us to factor \( x \) out of the entire equation, simplifying it greatly. This process transforms the original cubic equation into a product of a linear factor and a quadratic expression.
- Common Factor: Look for common terms in all parts of an equation. In this case, \( x \) was common in all terms, so it was factored out.
- Rewriting the Equation: After factoring, the equation became \( x(x^2 - 12x + 32) = 0 \), which separates the solution process for different factors.
Quadratic Formula
The quadratic formula provides a means to find the roots of any quadratic equation, expressed as \( ax^2 + bx + c = 0 \). It's a universal tool and especially handy when the quadratic equation does not factor neatly. The formula is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula directly takes the coefficients \( a \), \( b \), and \( c \) from the equation and provides the possible values of \( x \).
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula directly takes the coefficients \( a \), \( b \), and \( c \) from the equation and provides the possible values of \( x \).
- Simplifying Steps: For our example, the coefficients are \( a = 1 \), \( b = -12 \), and \( c = 32 \). Plugging into the formula simplifies the process of solving the quadratic part.
- Two Solutions: The formula accounts for both the plus and minus in \( \pm \), which leads to two potential results for \( x \), depending on the discriminant.
Discriminant
The discriminant is a part of the quadratic formula with significant implications on the nature of the solutions of a quadratic equation. It is represented by \( \Delta \) and is calculated as:
\[\Delta = b^2 - 4ac\]The discriminant helps in figuring out how many and what kind of solutions will be present:
\[\Delta = b^2 - 4ac\]The discriminant helps in figuring out how many and what kind of solutions will be present:
- Positive Discriminant: This means there are two distinct real solutions to the quadratic equation. In our example, \( \Delta = 16 \), which led to real solutions as \( 4 \) and \( 8 \).
- Zero Discriminant: This results in one real double root, implying the parabola touches the x-axis at one point.
- Negative Discriminant: Indicates there are no real solutions, and the roots are complex or imaginary.