Problem 46
Question
Solve each equation in Exercises \(39-54\) by completing the square. $$ x^{2}+6 x-5=0 $$
Step-by-Step Solution
Verified Answer
The solution to the equation \( x^{2} + 6x - 5 = 0 \) by completing the square method is \( x = -3 \pm \sqrt{14} \).
1Step 1: Arrange the quadratic equation in the right form
First, arrange the given equation in the form \( ax^2 + bx = c \). This can be achieved by moving constant term to the other side of the equation. So, the equation \( x^{2} + 6x - 5 = 0 \) becomes \( x^{2} + 6x = 5 \).
2Step 2: Write quadratic term in squared binomial form
The next step is to re-write the quadratic term so that it is a squared binomial. The identity \( (a+b)^2 = a^2 + 2ab + b^2 \) is used to do this. Here, \( a = x \) and \( 2ab = 6x \) (i.e. \( b = 3 \)). So, we can write \( x^{2} + 6x = (x+3)^{2} - 3^{2} = (x+3)^{2} - 9 \). However, we initially established equation \( x^{2} + 6x = 5 \), therefore, this can be re-written as \( (x+3)^{2} - 9 = 5 \).
3Step 3: Solve for x
The final step is to isolate \( x \). start by adding 9 to both sides of equation resulting in \( (x+3)^{2} = 14 \). Taking square root of both sides gives \( x + 3 = \pm \sqrt{14} \) i.e. \( x = -3 \pm \sqrt{14} \).
Key Concepts
Quadratic EquationsSolving Quadratic EquationsBinomial Squares
Quadratic Equations
Quadratic equations are mathematical expressions of the form ax2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. They are so named because 'quadra' comes from the Latin word for 'square', and the highest power of the variable x is two, which symbolizes a square of the side x. These equations are everywhere in algebra and are essential for understanding numerous aspects of mathematics and applied sciences.
Recognizing a quadratic equation is easy because it consists of a square term x2, a linear term x, and a constant term without variables. Graphically, these equations represent parabolas, and the solutions of the equation correspond to points where the parabola crosses the x-axis. These solutions are also termed as 'roots' of the equation.
While there are several methods to solve quadratic equations, one fundamental approach is 'completing the square', which allows us to simplify the equation by transforming it into a perfect square trinomial and then easily finding the values of x that satisfy the equation.
Recognizing a quadratic equation is easy because it consists of a square term x2, a linear term x, and a constant term without variables. Graphically, these equations represent parabolas, and the solutions of the equation correspond to points where the parabola crosses the x-axis. These solutions are also termed as 'roots' of the equation.
While there are several methods to solve quadratic equations, one fundamental approach is 'completing the square', which allows us to simplify the equation by transforming it into a perfect square trinomial and then easily finding the values of x that satisfy the equation.
Solving Quadratic Equations
Solving quadratic equations can be done through various methods, including factoring, using the quadratic formula, graphing, or completing the square. When factoring or the quadratic formula are cumbersome, or when we wish to derive the quadratic formula itself, we often use the technique of completing the square.
This method involves manipulating the equation to form a perfect square trinomial on one side, thereby making it easier to solve for the variable x. It's called 'completing the square' because we add a certain value to both sides of the equation that allows us to express it as a binomial squared.
To solve an equation like x2 + 6x - 5 = 0 by completing the square:
This method involves manipulating the equation to form a perfect square trinomial on one side, thereby making it easier to solve for the variable x. It's called 'completing the square' because we add a certain value to both sides of the equation that allows us to express it as a binomial squared.
To solve an equation like x2 + 6x - 5 = 0 by completing the square:
- First, move the constant term to the other side of the equation.
- Then find the number that makes the x-terms into a perfect square trinomial.
- After forming the perfect square on one side, take the square root of both sides to solve for x.
- Finally, simplify to determine the solutions or roots of the equation.
Binomial Squares
A binomial square is a type of algebraic expression that contains two terms and is squared. The general form is (a + b)2, which when expanded, becomes a2 + 2ab + b2. This identity is crucial for the process of completing the square since it helps us to rewrite quadratic equations into a format that can be more easily solved.
To transform the quadratic term into a squared binomial, as seen in the exercise x2 + 6x = 5, we use the middle term (which is 6x in this case) to determine our 'b' in the binomial (x + b)2. We then subtract the square of 'b' to balance the equation, making it a perfect square trinomial.
To transform the quadratic term into a squared binomial, as seen in the exercise x2 + 6x = 5, we use the middle term (which is 6x in this case) to determine our 'b' in the binomial (x + b)2. We then subtract the square of 'b' to balance the equation, making it a perfect square trinomial.
Expanded Binomial Example:
(x + 3)2 becomes x2 + 6x + 9 after application of the binomial square expansion. This technique not only simplifies the equation but also paves the way for the solution by taking square roots. Binomial squares are foundational in algebra and help bridge the gap between various forms of quadratic equations and their solutions.Other exercises in this chapter
Problem 46
Explain how to add complex numbers. Provide an example with your explanation.
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Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \fra
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Solve each linear inequality in Exercises 27-48 and graph the solution set on a number line. Express the solution set using interval notation. $$\frac{4 x-3}{6}
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Exercises \(31-50\) contain equations with variables in denominators. For each equation, a. Write the value or values of the variable that make a denominator ze
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