Chapter 10: Problem 33
Solve each of the following quadratic equations using the method that seems most appropriate to you. $$x+\frac{1}{x}=\frac{25}{12}$$
Short Answer
Expert verified
The solutions are \(x = \frac{4}{3}\) and \(x = \frac{3}{4}\).
Step by step solution
01
Clear the Fraction
The given equation is \(x + \frac{1}{x} = \frac{25}{12}\). To eliminate the fraction, multiply the entire equation by \(x\) to get rid of the denominator. This gives us:\[x^2 + 1 = \frac{25}{12}x\]
02
Rearrange the Equation
Next, let's bring all terms to one side of the equation to form a standard quadratic equation. So, subtract \(\frac{25}{12}x\) from both sides:\[x^2 - \frac{25}{12}x + 1 = 0\]
03
Clear Fractions by Multiplying
To make the equation easier to work with, multiply every term by 12 to remove fractions:\[12x^2 - 25x + 12 = 0\]
04
Calculate the Discriminant
Here, we have a quadratic equation in standard form: \(ax^2 + bx + c = 0\), where \(a = 12\), \(b = -25\), and \(c = 12\). The discriminant \(D\) is given by \(b^2 - 4ac\). Substitute the values:\[D = (-25)^2 - 4 \times 12 \times 12 = 625 - 576 = 49\]
05
Use the Quadratic Formula
Since the discriminant is positive, there are two real solutions. Use the quadratic formula:\[x = \frac{-b \pm \sqrt{D}}{2a}\]Substitute the values \(b = -25\), \(a = 12\), and \(D = 49\):\[x = \frac{25 \pm \sqrt{49}}{24}\]
06
Calculate the Solutions
Compute the solutions using the quadratic formula:1. \(x = \frac{25 + 7}{24} = \frac{32}{24} = \frac{4}{3}\)2. \(x = \frac{25 - 7}{24} = \frac{18}{24} = \frac{3}{4}\)These give the two solutions: \(x = \frac{4}{3}\) and \(x = \frac{3}{4}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
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 for finding the solutions of quadratic equations. When an equation is in the form of \( ax^2 + bx + c = 0 \), we can use the quadratic formula to calculate the roots of the equation. The roots are given by:
This method is reliable and especially useful when the quadratic expression doesn’t factorize easily. It saves time and ensures accuracy in solving complex equations.
- \( x = \frac{-b \pm \sqrt{D}}{2a} \)
This method is reliable and especially useful when the quadratic expression doesn’t factorize easily. It saves time and ensures accuracy in solving complex equations.
Discriminant
The discriminant \( D \) plays a crucial role in understanding the nature of the roots of a quadratic equation. It's calculated as \( D = b^2 - 4ac \), derived from the coefficients of the standard form \( ax^2 + bx + c = 0 \). Here are the implications:
This indicator helps plan the approach to solving the equation, as positive discriminants tell us to expect two different values for \( x \) when we solve using the quadratic formula.
- If \( D > 0 \), the equation has two distinct real solutions.
- If \( D = 0 \), there is exactly one real solution, or a double root.
- If \( D < 0 \), the solutions are complex or imaginary.
This indicator helps plan the approach to solving the equation, as positive discriminants tell us to expect two different values for \( x \) when we solve using the quadratic formula.
Standard Form of a Quadratic Equation
The standard form of a quadratic equation is expressed as \( ax^2 + bx + c = 0 \). This format is crucial as it sets up the equation for operations like factoring, completing the square, and using the quadratic formula. Each term in the equation:
This rearranging process is key as it prepares the equation for solution using various algebraic methods.
- \( ax^2 \): Quadratic term with \( a \) as the coefficient of \( x^2 \).
- \( bx \): Linear term with \( b \) as the coefficient of \( x \).
- \( c \): Constant term.
This rearranging process is key as it prepares the equation for solution using various algebraic methods.