Chapter 7: Problem 27
In Exercises \(11-30,\) solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$9 x^{2}+6 x+1=0$$
Short Answer
Expert verified
The solution to the equation is \( x = -\frac{1}{3} \).
Step by step solution
01
Write the Equation in Standard Form
The given equation is already in standard quadratic form: \( ax^2 + bx + c = 0 \), where \( a = 9 \), \( b = 6 \), and \( c = 1 \). The equation is \( 9x^2 + 6x + 1 = 0 \).
02
Divide by the Coefficient of x^2
To simplify completing the square, divide the entire equation by 9, the coefficient of \( x^2 \). This gives us: \( x^2 + \frac{2}{3}x + \frac{1}{9} = 0 \).
03
Move the Constant Term
Move the constant term \( \frac{1}{9} \) to the other side of the equation: \( x^2 + \frac{2}{3}x = -\frac{1}{9} \).
04
Complete the Square
To complete the square, take half of the coefficient of \( x \), which is \( \frac{2}{3} \), divide it by 2 to get \( \frac{1}{3} \), and then square it to get \( \frac{1}{9} \). Add \( \frac{1}{9} \) to both sides of the equation: \( x^2 + \frac{2}{3}x + \frac{1}{9} = 0 \). This step makes the left side a perfect square trinomial.
05
Rewrite as a Perfect Square
The left side of the equation \( x^2 + \frac{2}{3}x + \frac{1}{9} \) can be written as a perfect square: \((x + \frac{1}{3})^2 = 0 \).
06
Solve the Perfect Square Equation
Take the square root of both sides: \( x + \frac{1}{3} = 0 \).
07
Solve for x
Subtract \( \frac{1}{3} \) from both sides to isolate \( x \): \( x = -\frac{1}{3} \). This equation has one solution because it was a perfect square trinomial.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a technique used to solve quadratic equations by turning them into a perfect square trinomial. This method involves altering the equation so that it can be expressed as the square of a binomial. Here's how it works:
1. **Start with the standard form**: Begin with a quadratic equation in the standard form, \( ax^2 + bx + c = 0 \).
2. **Divide through by \(a\)**: If \(a\) is not 1, divide every term by \(a\) to simplify.
3. **Move the constant term**: Shift the constant \(c\) to the other side of the equation.
4. **Create a perfect square**: Take half of the coefficient of \(x\), square it, and add it to both sides of the equation. This step is crucial to forming a perfect square trinomial.
This method helps in finding the roots of the equation by converting it into a simpler form. Completing the square makes it easier to see the solution visually or to solve it algebraically.
1. **Start with the standard form**: Begin with a quadratic equation in the standard form, \( ax^2 + bx + c = 0 \).
2. **Divide through by \(a\)**: If \(a\) is not 1, divide every term by \(a\) to simplify.
3. **Move the constant term**: Shift the constant \(c\) to the other side of the equation.
4. **Create a perfect square**: Take half of the coefficient of \(x\), square it, and add it to both sides of the equation. This step is crucial to forming a perfect square trinomial.
This method helps in finding the roots of the equation by converting it into a simpler form. Completing the square makes it easier to see the solution visually or to solve it algebraically.
Perfect Square Trinomial
A perfect square trinomial is an expression that can be factored into the square of a binomial. These take the form \( (x + d)^2 \) or \( (x - d)^2 \). Recognizing a perfect square trinomial is key in solving quadratic equations through completing the square.
To create a perfect square trinomial:
In the original equation \( x^2 + \frac{2}{3}x = -\frac{1}{9} \), adding \( \frac{1}{9} \) to both sides turns it into \( (x + \frac{1}{3})^2 = 0 \). Now it's evident as a perfect square trinomial.
To create a perfect square trinomial:
- Take the coefficient of the linear term (that is \(b\) if you have \(x^2 + bx\)).
- Divide it by 2 and square the result.
In the original equation \( x^2 + \frac{2}{3}x = -\frac{1}{9} \), adding \( \frac{1}{9} \) to both sides turns it into \( (x + \frac{1}{3})^2 = 0 \). Now it's evident as a perfect square trinomial.
Factoring Quadratics
Factoring quadratics is a straightforward method for solving quadratic equations, typically when they can be easily broken down into simpler expressions. It involves expressing the quadratic in the form \( (px + q)(rx + s) = 0 \).
Here's a brief how-to guide:
Here's a brief how-to guide:
- Find two numbers that multiply to \(ac\) (the coefficient of \(x^2\) times the constant term) and add up to \(b\) (the coefficient of \(x\)).
- Split the middle term using these two numbers.
- Factor by grouping, effectively reducing the quadratic into two binomials.
Quadratic Formula
The quadratic formula is a foolproof method to solve any quadratic equation and is especially useful when other methods like factoring aren't applicable. It derives from the standard quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using the quadratic formula involves:
Whether or not a particular quadratic equation can be factored or not, the quadratic formula will always get you the solution. It's especially handy for solving quadratics where the roots are irrational or complex.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Using the quadratic formula involves:
- Identifying the coefficients \(a\), \(b\), and \(c\) in the equation \(ax^2 + bx + c = 0\).
- Plugging these values into the formula.
- Simplifying to find the solution for \(x\).
Whether or not a particular quadratic equation can be factored or not, the quadratic formula will always get you the solution. It's especially handy for solving quadratics where the roots are irrational or complex.