Problem 22 - College Algebra Essentials

Question

Solve each equation in Exercises \(15-34\) by the square root property. $$ (x-3)^{2}=36 $$

Step-by-Step Solution

Answer

The solutions to the equation \((x-3)^{2}=36\) are \(x = 9\) and \(x = -3\).

Step 1: Rearrange the equation

Firstly, The squared binomial expression \((x-3)^{2}\) is already isolated on one side of the equation, so no rearrangement is necessary. The given equation is \((x-3)^{2}=36.

Step 2: Apply the square root property

The square root property asserts that if \(p ^{2}=q\), then \(p = \sqrt{q}\) or \(p = -\sqrt{q}\). Applying this property to the equation gives two equations: \(x - 3 = \sqrt{36}\) and \(x - 3 = -\sqrt{36}\). Simplifying, we obtain \(x - 3 = 6\) and \(x - 3 = -6\).

Step 3: Solve for the variable

For each of the previous equations, solving for \(x\) involves adding 3 to both sides. Doing so results in \(x = 6 + 3 = 9\) and \(x = -6 + 3 = -3\).

Key Concepts

Square Root Property

The square root property is a valuable tool used to solve equations involving squares. When you have an equation in the form of

• \( p^{2} = q \),

you can apply this property. It tells us you have two possible solutions: \( p = \sqrt{q} \) and \( p = -\sqrt{q} \).
This is because squaring either a positive or a negative number will result in the same positive square.
This property is very handy when dealing with quadratic equations of the form \((x-a)^2 = b\). It can quickly lead us to the solutions without the need to expand the expression first.
By knowing both potential values for the square, you can find all solutions to the equation and hence, tackle quadratic equations efficiently.

Quadratic Solutions

Quadratic solutions are the values for which a quadratic equation is satisfied. A quadratic equation can have zero, one, or two solutions, depending on its discriminant and the method applied.
The square root property simplifies finding these solutions when the quadratic is in the form of a perfect square, like \((x-3)^{2} = 36\).
This setup provides a direct path:

• Isolate the squared term,
• Apply the square root property to both sides,
• Solve the resulting linear equations.

In the given example, after applying the square root property, we arrived at two linear equations: \(x - 3 = 6\) and \(x - 3 = -6\).
Solving these gives us the solutions \(x = 9\) and \(x = -3\). Quadratic solutions give insights into the behavior of parabolic curves, which are crucial in various applications.

Isolating Variables

Isolating variables is a fundamental technique in solving equations. It involves getting the variable on one side of the equation to find its value.
In the quadratic example \((x-3)^{2} = 36\), the expression is already isolated, allowing for straightforward solving.
Once you apply the square root property, you get equations like \(x - 3 = 6\) and \(x - 3 = -6\).
To isolate \(x\), you simply perform basic arithmetic:

• Add 3 to both sides of each equation.

By doing so, you determine the values of \(x\) as \(x = 9\) and \(x = -3\).
Isolating variables is crucial because it helps solve for unknowns in various mathematical contexts efficiently.