Chapter 7: Problem 19
In Exercises \(11-30,\) solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$2 s^{2}+5 s=3$$
Short Answer
Expert verified
The solutions are \(s = \frac{1}{2}\) and \(s = -3\).
Step by step solution
01
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:\[ 2s^2 + 5s - 3 = 0 \]
02
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:\[ 2s^2 + 5s = 3 \]
03
Divide by Leading Coefficient
To complete the square, the quadratic term must have a coefficient of 1. Divide the entire equation by 2:\[ s^2 + \frac{5}{2}s = \frac{3}{2} \]
04
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:\[ \left(s + \frac{5}{4}\right)^2 = \frac{3}{2} + \frac{25}{16} \]
05
Simplify Right Side
Find a common denominator to simplify the right side:\[ \frac{3}{2} = \frac{24}{16} \]\[ \frac{3}{2} + \frac{25}{16} = \frac{24}{16} + \frac{25}{16} = \frac{49}{16} \]
06
Solve for the Variable
Take the square root of both sides and solve for \(s\):\[ s + \frac{5}{4} = \pm \sqrt{\frac{49}{16}} \]\[ s + \frac{5}{4} = \pm \frac{7}{4} \]
07
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:\[ s = \frac{2}{4} = \frac{1}{2} \]\[ s = -\frac{12}{4} = -3 \]
08
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:\[ 2s - 1 = 0 \Rightarrow s = \frac{1}{2} \]\[ s + 3 = 0 \Rightarrow s = -3 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
- Completing the square
- Factoring
- Using the quadratic formula
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:
- \( mx + n = 0 \)
- \( px + q = 0 \)
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.