Problem 19 - Basic Technical Mathematics with Calculus

Question

In Exercises \(11-30,\) solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$2 s^{2}+5 s=3$$

Step-by-Step Solution

Answer

The solutions are \(s = \frac{1}{2}\) and \(s = -3\).

Step 1: Rewrite Equation in Standard Form

Begin by moving all terms to one side of the equation to get it in standard form. Subtract 3 from both sides:\[ 2s^2 + 5s - 3 = 0 \]

Step 2: Isolate the Quadratic and Linear Terms

Next, we want to isolate the terms involving \(s\). Start by moving the constant term to the other side:\[ 2s^2 + 5s = 3 \]

Step 3: Divide by Leading Coefficient

To complete the square, the quadratic term must have a coefficient of 1. Divide the entire equation by 2:\[ s^2 + \frac{5}{2}s = \frac{3}{2} \]

Step 4: Complete the Square

To complete the square, add and subtract the square of half the coefficient of \(s\). The coefficient of \(s\) is \(\frac{5}{2}\), so take half of it, \(\frac{5}{4}\), and square it to get \(\frac{25}{16}\):\[ s^2 + \frac{5}{2}s + \left(\frac{5}{4}\right)^2 = \frac{3}{2} + \left(\frac{5}{4}\right)^2 \]Simplify and write as a perfect square:\[ \left(s + \frac{5}{4}\right)^2 = \frac{3}{2} + \frac{25}{16} \]

Step 5: Simplify Right Side

Find a common denominator to simplify the right side:\[ \frac{3}{2} = \frac{24}{16} \]\[ \frac{3}{2} + \frac{25}{16} = \frac{24}{16} + \frac{25}{16} = \frac{49}{16} \]

Step 6: Solve for the Variable

Take the square root of both sides and solve for \(s\):\[ s + \frac{5}{4} = \pm \sqrt{\frac{49}{16}} \]\[ s + \frac{5}{4} = \pm \frac{7}{4} \]

Step 7: Isolate and Find Solutions

Isolate \(s\) by subtracting \(\frac{5}{4}\):\[ s = -\frac{5}{4} + \frac{7}{4} \quad \text{or} \quad s = -\frac{5}{4} - \frac{7}{4} \]Solve each equation:\[ s = \frac{2}{4} = \frac{1}{2} \]\[ s = -\frac{12}{4} = -3 \]

Step 8: Verify by Factoring (Optional)

Factor the original quadratic equation to verify the solutions:The equation \( 2s^2 + 5s - 3 = 0 \) factors as:\[ (2s - 1)(s + 3) = 0 \]The factors give the solutions:\[ 2s - 1 = 0 \Rightarrow s = \frac{1}{2} \]\[ s + 3 = 0 \Rightarrow s = -3 \]

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental type of polynomial equation in algebra. They are characterized by their highest degree term being squared, hence the name "quadratic." A standard quadratic equation can be expressed in the form:\[ ax^2 + bx + c = 0 \]where:

• \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \) since it is the leading coefficient.
• \( x \) is the variable that represents an unknown value we aim to find.

Quadratic equations commonly appear in problems involving projectiles, areas, and optimization. Understanding how to manipulate these equations is essential for solving them effectively. The solutions can be real or complex, and can sometimes be found using various methods, such as:

• Completing the square
• Factoring
• Using the quadratic formula

Each method has its own advantages, and choosing one depends on the form of the quadratic equation and the context of the problem.

Explaining Factoring

Factoring is a method used to rewrite an equation as a product of simpler expressions, making it easier to find the roots. When a quadratic equation can be factored, it is often the simplest method to use to find the solutions. Generally, the standard form \( ax^2 + bx + c = 0 \) is rewritten into:\[ (mx + n)(px + q) = 0 \]By setting each factor to zero, the solutions can be obtained:

• \( mx + n = 0 \)
• \( px + q = 0 \)

These equations can then be solved to find the values of \( x \). Factoring requires that you identify products and sums that match the given quadratic equation. If successful, this method provides a straightforward way to find the roots willingly. Remember that not all quadratic equations are easily factored, which might make another method more practical in those cases.

Solving Quadratic Equations Using Completing the Square

Completing the square is a method for solving quadratic equations by transforming a quadratic into a perfect square trinomial. This technique is especially useful when the equation cannot be easily factored. Here's the step-by-step process:

• First, ensure the equation is in the form \( ax^2 + bx = -c \) and isolate the quadratic and linear terms from the constant.
• Make the coefficient of \( x^2 \) equal to 1 by dividing the equation by \( a \), if necessary.
• Add and subtract the square of half the coefficient of \( x \) to both sides of the equation to complete the square.
• This transforms the left side of the equation into a perfect square trinomial, looking like \((x + d)^2\).
• Once in this form, take the square root of both sides to solve for \( x \).
• Simplify to find the final solutions.

This method may initially seem lengthy, but it provides a systematic approach to finding exact solutions even when other methods may not apply, such as when dealing with irrational numbers or certain types of word problems.