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Magazine Chapter 7: Problem 29
Solve the given quadratic equations by completing the square. $$9 x^{2}+6 x+1=0$$
Short Answer
Expert verified
The solution is x = -1/3.
Step by step solution
01
Ensure the Equation is in Standard Quadratic Form
First, we observe that the given quadratic equation is already in the standard quadratic form: \(9x^2 + 6x + 1 = 0\). The terms are arranged by descending powers of \(x\).
02
Divide by the Coefficient of x虏
The leading coefficient in the equation is 9. To simplify completing the square, divide each term by 9:\[ x^2 + \frac{2}{3}x + \frac{1}{9} = 0 \]
03
Move the Constant to the Other Side
Subtract \(\frac{1}{9}\) from both sides to isolate the terms involving \(x\):\[ x^2 + \frac{2}{3}x = -\frac{1}{9} \]
04
Complete the Square
To complete the square, add \[\left(\frac{1}{2} \cdot \frac{2}{3}\right)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]to both sides. The equation becomes:\[ x^2 + \frac{2}{3}x + \frac{1}{9} = -\frac{1}{9} + \frac{1}{9} \] \[ x^2 + \frac{2}{3}x + \frac{1}{9} = 0 \]
05
Rewrite as a Perfect Square
The left side is now a perfect square trinomial. Rewrite it as a square:\[ \left(x + \frac{1}{3}\right)^2 = 0 \]
06
Solve for x
Take the square root of both sides:\[ x + \frac{1}{3} = 0 \]Subtract \(\frac{1}{3}\) from both sides to find \(x\):\[ x = -\frac{1}{3} \]
07
Conclusion
So the solution to the equation \(9x^2 + 6x + 1 = 0\) by completing the square is \(x = -\frac{1}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
When solving quadratic equations, completing the square is a technique used to transform a quadratic expression into a perfect square trinomial. This method simplifies finding solutions by rewriting the quadratic in a form that can be easily solved by taking the square root. Here's how it's done:
- Identify the given quadratic equation. Make sure it's in the form \( ax^2 + bx + c = 0 \), then divide all terms by \( a \) to simplify the equation to \( x^2 + \frac{b}{a}x = -\frac{c}{a} \).
- The goal is to add a term to complete the square. This is calculated as \( \left(\frac{b}{2a}\right)^2 \).
- Add this term to both sides of the equation to keep it balanced.
- The left side can now be factored into a binomial squared.
Quadratic Formula
The quadratic formula provides a direct way to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). It's a universal method used when completing the square is cumbersome or when the equation has complicated numbers. The formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it works:
- Calculate the determinant \( b^2 - 4ac \). This part of the formula helps determine the nature of the roots (real or complex).
- Substitute the values of \( a \), \( b \), and \( c \) directly into the formula.
- Solve for \( x \) to get the possible solutions for the equation.
Standard Form
The standard form of a quadratic equation is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). Here's why it's important:
- This form makes it easier to apply methods like factoring, completing the square, or using the quadratic formula to find solutions.
- Arranging an equation in standard form allows for consistent analysis and comparison between different quadratic equations.
- It highlights the coefficients and constants, offering better insights when graphing quadratic functions as parabolas.
Perfect Square Trinomial
A perfect square trinomial is a special type of quadratic expression that can be expressed as the square of a binomial. For instance, the expression \( (x + \frac{1}{3})^2 \) results in a perfect square trinomial when expanded: \( x^2 + \frac{2}{3}x + \frac{1}{9} \). It's identified through:
- Checking if the middle term is twice the product of the square root of the first and last terms.
- Knowing that perfect square trinomials make completing the square technique straightforward since they are already in squared binomial form.
- Having the ability to factor it easily, knowing that it results in a single repeated binomial squared.
