Problem 72
Question
Solve for \(x .\) Assume that a and \(b\) represent positive real numbers. \(9 x^{2}-25 a=0\)
Step-by-Step Solution
Verified Answer
x = \frac{5\text{sqrt}(a)}{3} or x = -\frac{5\text{sqrt}(a)}{3}
1Step 1: Rewrite the equation
Start by rewriting the given equation so that the term containing the variable is isolated. The given equation is:\[\begin{equation}9x^2 - 25a = 0ewline\text{Add } 25a \text{ to both sides to get:}ewline9x^2 = 25aewline\end{equation}\]
2Step 2: Solve for the variable term
Next, divide both sides of the equation by 9 to isolate the squared term involving the variable:\[\begin{equation}x^2 = \frac{25a}{9}ewline\end{equation}\]
3Step 3: Take the square root of both sides
Take the square root of both sides of the equation to solve for x. Remember to consider both the positive and negative roots:\[\begin{equation}x = \frac{\text{sqrt}(25a)}{\text{sqrt}(9)} = \frac{5\text{sqrt}(a)}{3}ewline\text{Thus,}ewlinex = \frac{5\text{sqrt}(a)}{3} \text{ or } x = -\frac{5\text{sqrt}(a)}{3}ewline\end{equation}\]
Key Concepts
Isolating VariablesDividing Both Sides of an EquationSquare Roots in Algebra
Isolating Variables
Isolating the variable is a critical first step when solving equations. This simply means getting the variable we want to solve for on one side of the equation by itself.
In this exercise, the variable is already surrounded by other terms. So, we rearrange the equation to isolate it.
Given the equation, \(9x^2 -25a = 0\), we want to move the term \(-25a\) to the other side where the constants are.
Start by adding \(25a\) to both sides to get: \(9x^2 = 25a\).
Now, we have isolated the term with the variable, \(9x^2\).
Consistency is key when isolating variables, always ensure the equation remains balanced by performing the same operation on both sides.
In this exercise, the variable is already surrounded by other terms. So, we rearrange the equation to isolate it.
Given the equation, \(9x^2 -25a = 0\), we want to move the term \(-25a\) to the other side where the constants are.
Start by adding \(25a\) to both sides to get: \(9x^2 = 25a\).
Now, we have isolated the term with the variable, \(9x^2\).
Consistency is key when isolating variables, always ensure the equation remains balanced by performing the same operation on both sides.
Dividing Both Sides of an Equation
Dividing both sides of an equation is a fundamental operation used to simplify and solve equations.
After isolating the variable term in the equation \(9x^2 = 25a\), our goal is to get \(x^2\) by itself.
To achieve that, divide both sides of the equation by 9, which is the coefficient of \(x^2\).
This yields: \[ x^2 = \frac{25a}{9} \].
By dividing both sides by the same amount, the equation remains balanced. This step simplifies the equation and brings us closer to the solution.
After isolating the variable term in the equation \(9x^2 = 25a\), our goal is to get \(x^2\) by itself.
To achieve that, divide both sides of the equation by 9, which is the coefficient of \(x^2\).
This yields: \[ x^2 = \frac{25a}{9} \].
By dividing both sides by the same amount, the equation remains balanced. This step simplifies the equation and brings us closer to the solution.
Square Roots in Algebra
Taking the square root of both sides of an equation is the final step in solving quadratic equations.
Given \(x^2 = \frac{25a}{9}\), we need to find the value of \(x\).
Taking the square root of both sides, we get: \[ x = \frac{\text{sqrt}(25a)}{\text{sqrt}(9)} = \frac{5\text{sqrt}(a)}{3} \]
Remember to consider both the positive and negative roots: \[ x = \frac{5\text{sqrt}(a)}{3} \text{ or } x = -\frac{5\text{sqrt}(a)}{3} \].
In algebra, squaring and taking the square root are inverse operations, so they cancel each other out.
This step provides the final solution. Always check for both positive and negative roots as these can sometimes be overlooked.
Given \(x^2 = \frac{25a}{9}\), we need to find the value of \(x\).
Taking the square root of both sides, we get: \[ x = \frac{\text{sqrt}(25a)}{\text{sqrt}(9)} = \frac{5\text{sqrt}(a)}{3} \]
Remember to consider both the positive and negative roots: \[ x = \frac{5\text{sqrt}(a)}{3} \text{ or } x = -\frac{5\text{sqrt}(a)}{3} \].
In algebra, squaring and taking the square root are inverse operations, so they cancel each other out.
This step provides the final solution. Always check for both positive and negative roots as these can sometimes be overlooked.
Other exercises in this chapter
Problem 70
Solve each equation. Check the solutions. \(3-2(x-1)^{-1}=(x-1)^{-2}\)
View solution Problem 71
Solve using the square root property. Simplify all radicals. $$ (4 x-1)^{2}-48=0 $$
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Solve using the square root property. Simplify all radicals. $$ (2 x-5)^{2}-180=0 $$
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The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(\left(x-\frac{1}{2}\right)^{
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