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Magazine Chapter 5: Problem 46
Let \(f(x)=x^{3}-2 x^{2}-8 x+10 .\) For what value(s) of \(x\) is \(f(x)=10 ?\)
Short Answer
Expert verified
The value(s) of \(x\) for which \(f(x) = 10\) are \(x = 0\), \(x = 4\), and \(x = -2\).
Step by step solution
01
Understand the Equation
We are given the function \(f(x) = x^3 - 2x^2 - 8x + 10\) and need to find the value(s) of \(x\) for which \(f(x) = 10\). This means setting the equation \(x^3 - 2x^2 - 8x + 10 = 10\) and solving for \(x\).
02
Simplify the Equation
Set the equation \(x^3 - 2x^2 - 8x + 10 = 10\). Subtract 10 from both sides to simplify: \[ x^3 - 2x^2 - 8x + 10 - 10 = 0 \] which reduces to \[ x^3 - 2x^2 - 8x = 0 \].
03
Factor Out the Common Term
Notice that \(x\) is a common term in each part of the expression. Factor \(x\) out of the equation: \[ x(x^2 - 2x - 8) = 0 \].
04
Solve for Zero 'x'
The factored equation \(x(x^2 - 2x - 8) = 0\) gives us two solutions. First, set the first term \(x = 0\). This is one solution.
05
Solve the Quadratic Equation
Solve the quadratic part \(x^2 - 2x - 8 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = -2, c = -8\).
06
Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac\): \((-2)^2 - 4(1)(-8) = 4 + 32 = 36\). Since the discriminant is positive, there are two real solutions.
07
Apply the Quadratic Formula
Use the quadratic formula to find \(x\): \[ x = \frac{-(-2) \pm \sqrt{36}}{2(1)} = \frac{2 \pm 6}{2} \]. Thus, the solutions are \(x = \frac{2 + 6}{2} = 4\) and \(x = \frac{2 - 6}{2} = -2\).
08
List All Solutions
Combine the solutions from Steps 4 and 7: \(x = 0\), \(x = 4\), and \(x = -2\) are the values for which \(f(x) = 10\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of a quadratic equation. A quadratic equation is generally written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients. The formula to find the solutions for \(x\) is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula helps us determine both real and complex solutions to a quadratic equation. The symbol \(\pm\) indicates that there are typically two solutions: one for addition and one for subtraction. This means we often get two values for \(x\). When you're faced with any quadratic, you can rely on the quadratic formula as a reliable method to calculate its roots.
Factoring Polynomials
Factoring polynomials involves breaking down a polynomial expression into simpler terms, called factors, that can be multiplied together to form the original polynomial. In the example given, we factor out \(x\) from the equation \(x^3 - 2x^2 - 8x = 0\) to get \(x(x^2 - 2x - 8) = 0\).
- The first factor, \(x\), gives us an immediate solution, \(x = 0\).
- The second factor, \(x^2 - 2x - 8\), is a quadratic that can be solved using methods such as factoring further, completing the square, or using the quadratic formula.
Discriminant Calculation
The discriminant of a quadratic equation, given by \(b^2 - 4ac\), is a key factor in determining the nature of the solutions. Calculating the discriminant helps to predict how many real solutions exist for the equation.
- If the discriminant is positive, there are two distinct real solutions.
- If the discriminant is zero, there is exactly one real solution, which is also called a repeated or double root.
- If the discriminant is negative, there are no real solutions; instead, the solutions are complex or imaginary.
Real Solutions
Real solutions are values of \(x\) that satisfy the equation and result in real numbers, not complex ones. In the scenario where a function such as \(f(x)\) equals zero, finding all real solutions involves analyzing the roots of the polynomial.
- For example, from the equation \(x(x^2 - 2x - 8) = 0\), we found that \(x = 0\) is one real solution directly by setting the factor \(x = 0\).
- Then, using the quadratic formula for \(x^2 - 2x - 8 = 0\), we calculated two additional real solutions: \(x = 4\) and \(x = -2\).
