Problem 15
Question
Solve each equation in Exercises \(15-34\) by the square root property. $$ 3 x^{2}=27 $$
Step-by-Step Solution
Verified Answer
The solution is \(x = \pm3\).
1Step 1: Isolate x-squared
First we must isolate \(x^2\) on one side of the equation. To do this, divide both sides of the equation by 3: \[x^2 = 27/3\] simplifies to \[x^2 = 9\].
2Step 2: Take the square root of both sides
Taking the square root of both sides gives the solutions for x. Remember that when taking square root, it could be positive or negative. Like so: \[\sqrt{x^2} = \pm{\sqrt{9}}\]. This simplifies to \(x = \pm3\).
Key Concepts
Quadratic EquationsIsolating VariablesPositive and Negative SolutionsSolving Equations
Quadratic Equations
Quadratic equations are expressions where the highest power of the variable is two. They often appear in the form \(ax^2 + bx + c = 0\). However, in this exercise, the equation has been simplified to a form that you can tackle using the square root property. Understanding quadratic equations helps you see why isolating terms and finding their roots is so powerful.
- Quadratic equations can have up to two solutions because of the squared variable.
- These solutions are known as the roots of the equation.
- By recognizing these equations, you can apply specific strategies like the square root property to solve them.
Isolating Variables
The first step in solving an equation by the square root property involves isolating the variable that is squared. This means getting the squared term, \(x^2\), by itself on one side of the equation. In our example, the equation begins as \(3x^2 = 27\).
- You perform isolation by dividing both sides of the equation by 3, resulting in \(x^2 = 9\).
- Isolating variables makes further operations, like taking square roots, straightforward.
- Sticking to this initial step helps you maintain clarity and prevents mistakes.
Positive and Negative Solutions
After isolating the squared variable, the next step involves dealing with the square root. It's crucial to remember that squaring involves both positive and negative numbers. For example, both \(3^2\) and \((-3)^2\) equal 9, which points to two solutions.
- When taking the square root, always consider both positive and negative possibilities.
- This is represented by the symbol \(\pm\), meaning "+ or -".
- In this exercise, \(x = \pm 3\) provides two solutions: \(x = 3\) and \(x = -3\).
Solving Equations
Solving an equation using the square root property simplifies the process of finding the variable's values. Once the squared variable is isolated and you take its square root, you arrive at the solutions. In this exercise, solving \(x^2 = 9\) led to \(x = \pm 3\).
- This method is quick because it directly finds solutions without additional factoring or completing the square.
- Remember to verify your solutions by substituting them back into the original equation.
- This verification ensures that the solutions satisfy the initial problem.
Other exercises in this chapter
Problem 15
In Exercises \(9-20,\) find each product and write the result in standard form. $$(3+5 i)(3-5 i)$$
View solution Problem 15
Find all values of \(x\) satisfying the given conditions. \(y_{1}=2 x+6, y_{2}=x+8, y_{3}=x,\) and the difference between 3 times \(y_{1}\) and 5 times \(y_{2}\
View solution Problem 15
Check all proposed solutions. $$ \sqrt{2 x+13}=x+7 $$
View solution Problem 15
Solve and check linear equation. \(25-[2+5 y-3(y+2)]=\) \(-3(2 y-5)-[5(y-1)-3 y+3]\)
View solution