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Magazine Chapter 7: Problem 3
Solve the equation \(2+\frac{5}{x}=\frac{12}{x^{2}}\) both algebraically and graphically, then compare your solutions.
Short Answer
Expert verified
The solutions to the equation are \(x = 1.5\) and \(x = -4\).
Step by step solution
01
Write the original equation
We start with the equation given in the problem: \[ 2 + \frac{5}{x} = \frac{12}{x^2} \]
02
Eliminate fractions
To eliminate the fractions, multiply every term by \(x^2\) which is the least common multiple of the denominators:\[ x^2 \cdot 2 + x^2 \cdot \frac{5}{x} = x^2 \cdot \frac{12}{x^2} \]This becomes:\[ 2x^2 + 5x = 12 \]
03
Rearrange to form a quadratic equation
Bring all terms to one side of the equation to form a standard quadratic equation:\[ 2x^2 + 5x - 12 = 0 \]
04
Solve the quadratic equation
We will use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 5\), and \(c = -12\).First, calculate the discriminant:\[ b^2 - 4ac = 25 - 4 \cdot 2 \cdot (-12) = 25 + 96 = 121 \]Now, substitute into the quadratic formula:\[ x = \frac{-5 \pm \sqrt{121}}{4} \]\[ x = \frac{-5 \pm 11}{4} \]This gives us two solutions:\[ x = \frac{6}{4} = 1.5 \]\[ x = \frac{-16}{4} = -4 \]
05
Graphical analysis
Consider the original equation \(2 + \frac{5}{x} = \frac{12}{x^2}\) and graph the functions \(y_1 = 2 + \frac{5}{x}\) and \(y_2 = \frac{12}{x^2}\).The points where the graphs intersect are the solutions to the equation. These should match the algebraic solutions \(x = 1.5\) and \(x = -4\).
06
Comparison of solutions
Both the algebraic and graphical approach result in the same solutions: \(x = 1.5\) and \(x = -4\). This consistency verifies our solutions are correct.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Algebraic Solution
To solve a quadratic equation algebraically, we start by transforming the given equation, so that it can be handled more easily. In this exercise, we have the equation: \[ 2 + \frac{5}{x} = \frac{12}{x^2} \]The goal is to eliminate the fractions. This is typically done by multiplying every term by the least common multiple of the denominators, which, in this case, is \(x^2\). Once the fractions are cleared, you have a standard quadratic equation of the form \(ax^2 + bx + c = 0\).
- Write the equation as \(2x^2 + 5x - 12 = 0\)
- The standard form allows us to use various methods to find the solution, such as factoring, completing the square, or the quadratic formula.
Graphical Solution
Solving a quadratic equation graphically involves plotting the functions on a coordinate graph and finding their intersection points. Here, the initial equation is set up as two separate functions:
- \(y_1 = 2 + \frac{5}{x}\)
- \(y_2 = \frac{12}{x^2}\)
Quadratic Formula
The quadratic formula is a reliable method used to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our case, substituting \(a = 2\), \(b = 5\), and \(c = -12\) into the formula, provides the solutions.
- First, calculate the discriminant \(b^2 - 4ac\). It's crucial because it determines the nature of the roots.
- Next, substitute into the quadratic formula and solve for \(x\).
Discriminant in Quadratics
The discriminant in a quadratic equation is found within the quadratic formula: \(b^2 - 4ac\). It offers significant insight into the nature of the equation's roots:
- If the discriminant is positive, the quadratic equation has two distinct real roots.
- If it is zero, there is one real root, meaning the parabola touches the x-axis.
- If negative, the roots are complex, indicating the parabola does not intersect the x-axis at all.
