Problem 39 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(31-44\) : Solve by using the quadratic formula. $$ \frac{5}{w^{2}}-\frac{10}{w}+2=0 $$

Step-by-Step Solution

Answer

The solutions for \( w \) are \( \frac{5}{5+\sqrt{15}} \) and \( \frac{5}{5-\sqrt{15}} \).

Step 1: Identify the quadratic equation

The given equation is \( \frac{5}{w^2} - \frac{10}{w} + 2 = 0 \). Recognize this as a quadratic equation in terms of \( \frac{1}{w} \), so let \( x = \frac{1}{w} \). This transforms the equation into \( 5x^2 - 10x + 2 = 0 \).

Step 2: Use the quadratic formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 5 \), \( b = -10 \), and \( c = 2 \). Substitute these values into the formula to find \( x \).

Step 3: Calculate the discriminant

The discriminant \( b^2 - 4ac \) determines the nature of the roots. Compute it as follows: \( (-10)^2 - 4 \times 5 \times 2 = 100 - 40 = 60 \).

Step 4: Solve for \( x \) using the formula

Plug the values into the quadratic formula: \( x = \frac{-(-10) \pm \sqrt{60}}{2 \times 5} = \frac{10 \pm \sqrt{60}}{10} \).

Step 5: Simplify and express the solution

Simplify \( \sqrt{60} \) to \( \sqrt{4 \times 15} = 2\sqrt{15} \). Thus, the solutions for \( x \) are \( x = \frac{10 + 2\sqrt{15}}{10} \) and \( x = \frac{10 - 2\sqrt{15}}{10} \). Further simplify to get \( x = 1 + \frac{\sqrt{15}}{5} \) and \( x = 1 - \frac{\sqrt{15}}{5} \).

Step 6: Convert back to \( w \)

Since \( x = \frac{1}{w} \), invert the values to solve for \( w \). This gives \( w = \frac{5}{5+\sqrt{15}} \) and \( w = \frac{5}{5-\sqrt{15}} \). Rationalize these denominators.

Key Concepts

Quadratic Formula

The quadratic formula is a staple tool for solving quadratic equations. These are equations that fit the format \( ax^2 + bx + c = 0 \). The formula is written as: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] This formula allows you to solve for \( x \) by simply plugging in the values for \( a \), \( b \), and \( c \).

• \( a \) is the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

In the exercise, we recognized the quadratic form by letting \( x = \frac{1}{w} \) and transforming the given equation into \( 5x^2 - 10x + 2 = 0 \). Plugging \( a = 5 \), \( b = -10 \), and \( c = 2 \) into our quadratic formula will allow us to find \( x \). Once you find \( x \), then transforming it back to \( w \) is a simple inversion of \( x \).

Discriminant

The discriminant is a component of the quadratic formula that indicates the nature of the roots of a quadratic equation. It is found in the expression beneath the square root in the quadratic formula: \( b^2 - 4ac \).

• 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 negative, the roots are complex (not real).

In this exercise, our discriminant was calculated as 60, a positive number. This means our equation has two real and distinct roots. The calculation was straightforward: \((-10)^2 - 4 \times 5 \times 2 = 60\). Keep in mind that understanding the discriminant helps predict and understand the nature of roots without full calculations.

Simplifying Square Roots

Simplifying square roots is an essential skill when using the quadratic formula. This occurs when the discriminant is not a perfect square, which means it's not a whole number. For instance, in the given exercise, we encountered \( \sqrt{60} \). Simplify it by factoring it into perfect squares, like so: \[ \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15} \] Here, \( 4 \) is a perfect square, allowing us to simplify \( \sqrt{60} \) to \( 2\sqrt{15} \). This simplification reduces the complexity of the equation and helps in further calculations. It allows us to express our results more elegantly, focusing just on irreducible roots.

Rationalizing Denominators

Rationalizing denominators is often needed when dealing with fractions that contain square roots in the denominator. The goal here is to eliminate these square roots to make the expression easier to handle and more elegant. In our case, after transforming back to \( w \), we had denominators such as \( \frac{1}{5 + \sqrt{15}} \) and \( \frac{1}{5 - \sqrt{15}} \). To rationalize these, multiply both the numerator and the denominator by the conjugate of the denominator: \[\frac{5}{5+\sqrt{15}} \times \frac{5-\sqrt{15}}{5-\sqrt{15}} \] The expression simplifies to reduce the square roots: \[\frac{5(5-\sqrt{15})}{(5+\sqrt{15})(5-\sqrt{15})} = \frac{5(5-\sqrt{15})}{25-15} = \frac{5(5-\sqrt{15})}{10}\] Rationalizing makes our solutions cleaner and easier to interpret, further helping to verify results and ensuring they are presented clearly.