Problem 50
Question
Solve equation by completing the square. $$ x^{2}+4 x=12 $$
Step-by-Step Solution
Verified Answer
The solutions to the equation \(x^{2}+4 x=12\) are \(x = 2\) and \(x = -6\).
1Step 1: Rewrite the equation to establish a perfect square
Start by rewriting the equation as: \(x^{2}+4x-12=0\). This is a quadratic equation of the form \(ax^{2}+bx+c=0\). Here, \(a=1\), \(b=4\) and \(c=-12\).
2Step 2: Complete the square
In the equation \(x^{2}+4x-12=0\), the 'b' term is 4. Half of 'b' is 2, so by completing the square, we rewrite the equation as \((x+2)^{2} -4 = 12\). This equation is obtained by adding and subtracting \((b/2)^{2}\) which equals \(2^{2} = 4\) to the equation \(x^{2}+4x-12\), and rearranging the terms to create a perfect square trinomial.
3Step 3: Solving for x
We add 4 to both sides of the equation to get \((x+2)^{2} = 16\). We take the square root of both sides and note that the square of a number can be positive or negative. This gives us \(x+2 = ±4\). Then we subtract 2 from both sides to solve for 'x', thus \(x = 2\) or \(x = -6\).
Key Concepts
Quadratic EquationPerfect Square TrinomialSolve for x
Quadratic Equation
When encountering a quadratic equation, our primary goal is to find the values of 'x' that make the equation true. Quadratic equations are always in the form of \(ax^2 + bx + c = 0\), where 'a', 'b', and 'c' are constants, and 'a' is not equal to zero. This form is pivotal as it allows for various methods of solving, such as factoring, using the quadratic formula, graphing, or completing the square, which is a technique useful in solving when the equation does not factor easily.
Completing the square is particularly powerful as it not only facilitates finding solutions but also aids in understanding the properties of the quadratic equation, such as the vertex of a parabola described by the equation. Understanding the quadratic equation's structure helps in exploring these properties and finding the solutions efficiently.
Completing the square is particularly powerful as it not only facilitates finding solutions but also aids in understanding the properties of the quadratic equation, such as the vertex of a parabola described by the equation. Understanding the quadratic equation's structure helps in exploring these properties and finding the solutions efficiently.
Perfect Square Trinomial
A perfect square trinomial occurs when a binomial is squared, resulting in the form of \((ax + b)^2 = a^2x^2 + 2abx + b^2\). The main characteristic of this is that the middle term is twice the product of the square root of the first and third term. In the context of solving quadratic equations, creating a perfect square on one side of the equation is a pivotal step for completing the square. This usually involves manipulating the equation to have a constant term that makes the quadratic and linear terms form a perfect square when added together.
For instance, given the quadratic equation \(x^2 + 4x = 12\), we can transform it into a perfect square trinomial by finding a number that, when added and subtracted (to maintain balance), completes the square for the x-terms. In this case, adding and subtracting \(4\) (which is \((2)^2\) and represents \((b/2)^2\) with 'b' being the linear coefficient from the original equation) does the trick, creating \((x+2)^2\) on one side. This method is not just a mechanical process but a strategic move that simplifies the equation and makes it much easier to solve.
For instance, given the quadratic equation \(x^2 + 4x = 12\), we can transform it into a perfect square trinomial by finding a number that, when added and subtracted (to maintain balance), completes the square for the x-terms. In this case, adding and subtracting \(4\) (which is \((2)^2\) and represents \((b/2)^2\) with 'b' being the linear coefficient from the original equation) does the trick, creating \((x+2)^2\) on one side. This method is not just a mechanical process but a strategic move that simplifies the equation and makes it much easier to solve.
Solve for x
The ultimate goal of solving a quadratic equation is to solve for x. This means isolating 'x' on one side of the equation to find its value(s). When an equation is a perfect square trinomial, this task becomes more straightforward. After completing the square and arriving at an equation such as \((x+m)^2 = n\), we employ the square root method, recognizing the principle that if \(a^2 = b\), then \(a = ±√b\).
This technique brings a crucial aspect to light: the presence of two possible solutions for 'x', one positive and one negative, because a squared number is always positive, whether the original was positive or negative. So, for the example \((x+2)^2 = 16\), taking the square root of both sides indicates that \(x+2\) could be both \(4\) and \(-4\). Therefore, resolving for 'x' provides us two possible solutions, \(x = 2 - 2\) or \(x = -4 - 2\), resulting in \(x = 2\) and \(x = -6\). This insight is invaluable for accurately solving quadratic equations in various contexts.
This technique brings a crucial aspect to light: the presence of two possible solutions for 'x', one positive and one negative, because a squared number is always positive, whether the original was positive or negative. So, for the example \((x+2)^2 = 16\), taking the square root of both sides indicates that \(x+2\) could be both \(4\) and \(-4\). Therefore, resolving for 'x' provides us two possible solutions, \(x = 2 - 2\) or \(x = -4 - 2\), resulting in \(x = 2\) and \(x = -6\). This insight is invaluable for accurately solving quadratic equations in various contexts.
Other exercises in this chapter
Problem 49
write each English sentence as an equation in two variables. Then graph the equation. The \(y\) -value is three decreased by the square of the \(x\) -value.
View solution Problem 50
In all exercises, other than \(\varnothing,\) use interval notation to express solution sets and graph each solution set on a number line. In Exercises \(27-50,
View solution Problem 50
Solve each equation in Exercises 41–60 by making an appropriate substitution. $$2 x^{\frac{2}{3}}+7 x^{\frac{1}{3}}-15=0$$
View solution Problem 50
Solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? \(S=P+P r t\) for \(t\)
View solution