Chapter 8: Problem 67
Use completing the square to solve each equation. See Example 10. $$ x^{2}+\frac{2}{3} x+7=0 $$
Short Answer
Expert verified
The solutions to the equation are \(x = -\frac{1}{3} \pm \frac{i\sqrt{62}}{3}\).
Step by step solution
01
Move Constant Term to the Right Side
Begin by rewriting the equation to isolate the constant term. Move the constant 7 to the right side of the equation by subtracting 7 from both sides.\[ x^2 + \frac{2}{3}x = -7 \]
02
Find the Coefficient to Complete the Square
To complete the square, take half of the coefficient of the linear term (\(\frac{2}{3}\)), then square it. \[ \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \] \[ \left(\frac{1}{3}\right)^2 = \frac{1}{9} \]Add and subtract \(\frac{1}{9}\) inside the equation.
03
Add and Subtract the Square Term
Introduce the square term to both sides of the equation. Add \(\frac{1}{9}\) on the left side and remember to adjust the equation on the right side as follows:\[ x^2 + \frac{2}{3}x + \frac{1}{9} = -7 + \frac{1}{9} \]
04
Simplify the Equation
Simplify the expression on the right side:\[ -7 + \frac{1}{9} = -\frac{63}{9} + \frac{1}{9} = -\frac{62}{9} \]
05
Rewrite as a Perfect Square
The left side is now a perfect square trinomial and can be factored as:\[ \left(x + \frac{1}{3}\right)^2 = -\frac{62}{9} \]
06
Solve for x by Taking the Square Root
Take the square root of both sides:\[ x + \frac{1}{3} = \pm \sqrt{-\frac{62}{9}} \]Since \(\sqrt{-1} = i\), this becomes:\[ x + \frac{1}{3} = \pm \frac{i\sqrt{62}}{3} \]
07
Isolate x
Subtract \(\frac{1}{3}\) from both sides to solve for \(x\):\[ x = -\frac{1}{3} \pm \frac{i\sqrt{62}}{3} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a polynomial equation of the second degree, usually written in the form \(ax^2 + bx + c = 0\). This standard form consists of three terms: the quadratic term \(ax^2\), the linear term \(bx\), and the constant term \(c\). When solving a quadratic equation, the goal is to find the value(s) of \(x\) that make the whole expression equal to zero.
There are various methods to solve quadratic equations, including:
There are various methods to solve quadratic equations, including:
- Factoring
- Using the quadratic formula
- Completing the square
Complex Numbers
Complex numbers arise when solving quadratic equations, particularly those that do not have real solutions. If, during the process of solving a quadratic via methods like completing the square, you encounter a negative number under a square root, you'll deal with complex numbers. Complex numbers are numbers in the form \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The imaginary unit \(i\) is defined by the property that \(i^2 = -1\). This means that \(\sqrt{-1} = i\).
For example, if you find \(x = \pm \sqrt{-\frac{62}{9}}\), you rewrite this as \(x = \pm \frac{i\sqrt{62}}{3}\). The presence of the \(i\) indicates the solution involves complex numbers. Understanding complex numbers is crucial for solving quadratic equations that don't intersect the x-axis on a real number plane.
For example, if you find \(x = \pm \sqrt{-\frac{62}{9}}\), you rewrite this as \(x = \pm \frac{i\sqrt{62}}{3}\). The presence of the \(i\) indicates the solution involves complex numbers. Understanding complex numbers is crucial for solving quadratic equations that don't intersect the x-axis on a real number plane.
Factoring Trinomials
Factoring trinomials is a key skill in algebra, often a starting step in solving quadratic equations. Trinomials are polynomials with three terms, usually expressed as \(ax^2 + bx + c\). To factor a trinomial:
- Look for two numbers that multiply to \(ac\) (the product of the quadratic and constant coefficients) and add to \(b\) (the linear coefficient).
- Rewrite the middle term using these two numbers to break it into two terms.
- Group the new expression into two binomials and factor them separately.