Problem 44 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$\frac{1}{x}+6=\frac{5}{\sqrt{x}}$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{1}{4} \) and \( x = \frac{1}{9} \).

Step 1: Establish a Common Denominator

To simplify the given equation \( \frac{1}{x} + 6 = \frac{5}{\sqrt{x}} \), start by multiplying all terms by the least common denominator, which is \( x \sqrt{x} \) to clear the fractions. This gives the equation \( \sqrt{x} + 6x\sqrt{x} = 5x \).

Step 2: Simplify the Equation

Distribute the multiplication across each term: \( \sqrt{x} + 6x\sqrt{x} = 5x \) becomes \( \sqrt{x} (1 + 6x) = 5x \).

Step 3: Isolate the Radical Term

To solve for \( x \), isolate the square root term on one side by dividing both sides of the equation by \( 1 + 6x \). This results in \( \sqrt{x} = \frac{5x}{1 + 6x} \).

Step 4: Square Both Sides

Square both sides of the equation to eliminate the square root. This gives \( x = \left( \frac{5x}{1 + 6x} \right)^2 \), resulting in \( x = \frac{25x^2}{(1 + 6x)^2} \).

Step 5: Clear the Fraction

Multiply both sides by \( (1 + 6x)^2 \) to clear the complex fraction, leading to \( x(1 + 6x)^2 = 25x^2 \).

Step 6: Expand and Simplify

Expand \( (1 + 6x)^2 \) to get \( 1 + 12x + 36x^2 \). Thus, the equation becomes \( x(1 + 12x + 36x^2) = 25x^2 \), simplifying to \( x + 12x^2 + 36x^3 = 25x^2 \).

Step 7: Rearrange the Equation

Move all terms to one side to form a polynomial: \( 36x^3 + 12x^2 + x - 25x^2 = 0 \). Combining like terms results in \( 36x^3 - 13x^2 + x = 0 \).

Step 8: Factor Out Common Terms

Factor out the common term \( x \) to simplify: \( x(36x^2 - 13x + 1) = 0 \). This gives two potential solutions: \( x = 0 \) or solving \( 36x^2 - 13x + 1 = 0 \).

Step 9: Solve the Quadratic

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve \( 36x^2 - 13x + 1 = 0 \). Here, \( a = 36 \), \( b = -13 \), and \( c = 1 \). The discriminant is \( (-13)^2 - 4 \times 36 \times 1 = 169 - 144 = 25 \).

Step 10: Evaluate the Roots

Calculate the roots using the quadratic formula: \( x = \frac{13 \pm 5}{72} \). This results in \( x = \frac{18}{72} = \frac{1}{4} \) and \( x = \frac{8}{72} = \frac{1}{9} \).

Step 11: Check for Valid Solutions

The original equation involves division by \( \sqrt{x} \), so \( x = 0 \) is not valid as it makes the original expression undefined. Thus, the valid solutions are \( \frac{1}{4} \) and \( \frac{1}{9} \).

Step 12: Express the Solution

The solutions of the equation \( \frac{1}{x} + 6 = \frac{5}{\sqrt{x}} \) can be expressed as the interval \( (\frac{1}{9}, \frac{1}{4}) \). However, since we are dealing with isolated points, the answer is the set \( \left\{ \frac{1}{4}, \frac{1}{9} \right\} \).

Key Concepts

Quadratic Formula

The quadratic formula is a mathematical tool used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a eq 0\). The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula allows you to calculate the two possible solutions (or roots) of a quadratic equation. Let's highlight some key points:

• The term under the square root, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots:
• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If it is negative, the roots are complex and not real.
• The "\(\pm\)" sign indicates the two potential solutions due to the square root.

In our exercise, solving \(36x^2 - 13x + 1 = 0\) with the quadratic formula helps us find the values of \(x\) where the polynomial equals zero.

Interval Notation

Interval notation is a concise way of writing the set of all real numbers between two endpoints. It's particularly useful in solutions to inequalities and can also express the domain and range of functions. An interval can include its endpoints (closed interval) or exclude them (open interval). Here's how it works:

• Open intervals: \((a, b)\) represents all numbers between \(a\) and \(b\), not including \(a\) or \(b\).
• Closed intervals: \([a, b]\) includes the endpoints, meaning \(a\) and \(b\) are part of the interval.
• Half-open intervals: \([a, b)\) or \((a, b]\), include one endpoint but not the other.

In the provided solution, while the equation is solved to identify points \(x = \frac{1}{4}\) and \(x = \frac{1}{9}\), they are expressed as isolated points rather than a continuous interval. Therefore, we denote the solution as a set: \(\left\{ \frac{1}{4}, \frac{1}{9} \right\}\).

Radicals

Radicals are expressions that involve roots, such as the square root or cube root. A common type is the square root, represented by \(\sqrt{x}\). Here’s a simple breakdown of some basic concepts:

• The square root of a number \(x\) is a value that, when multiplied by itself, gives \(x\).
• In equations, radicals often require both sides to be squared to eliminate the root for simplification.
• Simplifying radical expressions follows specific rules. For example, \(\sqrt{a^2} = a\) when \(a\) is non-negative.

Radicals appear in the original equation \(\frac{1}{x} + 6 = \frac{5}{\sqrt{x}}\), specifically in the term \(\sqrt{x}\). In solving, we isolated \(\sqrt{x}\) before squaring both sides, streamlining the equation to remove the radical and proceed with traditional algebraic methods.

Polynomial Equations

Polynomial equations are expressions set equal to zero and consist of variables and coefficients, with operations restricted to addition, subtraction, multiplication, and non-negative integer exponents. The equation solved here represents a third-degree polynomial:\[36x^3 - 13x^2 + x = 0\]Let's dig into the basics:

• A polynomial's degree is the highest power of its variable; here, it is 3, making it a cubic equation.
• Such equations can often be factored into products of simpler polynomials or forms, sometimes giving immediate solutions for specific values of \(x\).
• Factoring methods or the application of the quadratic formula (for lower-order parts) help determine solutions.

In the solution process, we factored out \(x\) to initially break down the equation, leading to a simplified quadratic component solved using the quadratic formula. Efficiently understanding polynomials and factorization is fundamental to solving higher-degree equations.