Problem 35 - Elementary Algebra

Question

For Problems \(33-50\), set up an equation and solve the problem. (Objective 2 ) $$ \text { A number is } \frac{21}{10} \text { larger than its reciprocal. Find the number. } $$

Step-by-Step Solution

Answer

The number is 2.5.

Step 1: Define the Variables

Let the number be represented by \( x \). Its reciprocal will then be \( \frac{1}{x} \).

Step 2: Set up the Equation

According to the problem, the number \( x \) is \( \frac{21}{10} \) larger than its reciprocal. This can be expressed as:\[ x = \frac{1}{x} + \frac{21}{10} \]

Step 3: Eliminate the Fraction

To eliminate the fraction, multiply every term by \( x \) (assuming \( x eq 0 \)) to obtain:\[ x^2 = 1 + \frac{21}{10}x \]

Step 4: Clear the Equation

Multiply each term by 10 to remove the fraction:\[ 10x^2 = 10 + 21x \]

Step 5: Rearrange into a Quadratic Equation

Rearrange the equation to bring all terms to one side:\[ 10x^2 - 21x - 10 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 10 \), \( b = -21 \), and \( c = -10 \):- Calculate the discriminant: \( b^2 - 4ac = 441 + 400 = 841 \).- Find the roots: \( x = \frac{21 \pm \sqrt{841}}{20} \).- Simplify to get \( x = \frac{21 \pm 29}{20} \): - \( x = \frac{50}{20} = 2.5 \) - \( x = \frac{-8}{20} = -0.4 \).

Step 7: Verify the Solutions

Check if both solutions satisfy the condition in the original problem:- For \( x = 2.5 \), \( \frac{1}{x} = \frac{1}{2.5} = 0.4 \), and \( 2.5 = 0.4 + \frac{21}{10} \) which is true.- For \( x = -0.4 \), \( \frac{1}{x} = -2.5 \), and \( -0.4 eq -2.5 + \frac{21}{10} \). So, this does not hold.

Key Concepts

Understanding Reciprocals

In mathematics, a reciprocal is essentially what you get when you exchange the numerator and the denominator of a fraction. For any non-zero number \( x \), the reciprocal is \( \frac{1}{x} \). This simple operation is quite significant because when you multiply a number by its reciprocal, the result is always 1. That's the defining property:

• For example, the reciprocal of 5 is \( \frac{1}{5} \), and \( 5 \times \frac{1}{5} = 1 \).
• Similarly, the reciprocal of \( \frac{2}{3} \) is \( \frac{3}{2} \), and \( \frac{2}{3} \times \frac{3}{2} = 1 \).

In the given exercise, the reciprocal of a number \( x \) is used to form an equation that expresses a particular relationship between \( x \) and its reciprocal.

The Role of the Discriminant

The discriminant is an integral part of the quadratic formula and plays a crucial role in determining the nature of the roots of a quadratic equation. For any quadratic equation in the form \( ax^2 + bx + c = 0 \), the discriminant is given by the expression \( b^2 - 4ac \). The value of this discriminant indicates:

• When \( b^2 - 4ac > 0 \), the quadratic equation has two distinct real roots.
• If \( b^2 - 4ac = 0 \), the equation has exactly one real root, known as a repeated or double root.
• And if \( b^2 - 4ac < 0 \), the equation has no real roots but two complex conjugate roots.

In our problem, the discriminant \( 841 \) (since it is positive) suggests there are two distinct real solutions for the quadratic equation formed, which we calculated to be \( 2.5 \) and \(-0.4\).

Utilizing the Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations and can be used for any quadratic equation in the form of \( ax^2 + bx + c = 0 \). It is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides the solutions for \( x \) by using the coefficients \( a \), \( b \), and \( c \). Here's how to apply it effectively:

• Plug the values for \( a \), \( b \), and \( c \) into the formula.
• Calculate the discriminant \( b^2 - 4ac \) to discover the number and type of solutions.
• Apply the solutions formula to solve for \( x \).

In our exercise, with \( a = 10 \), \( b = -21 \), and \( c = -10 \), using the quadratic formula yields the roots \( x = 2.5 \) and \( x = -0.4 \). Only \( 2.5 \) is a valid solution because it satisfies the condition of being \( \frac{21}{10} \) larger than its reciprocal.