Problem 21 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ x^{2}=81 $$

Step-by-Step Solution

Answer

x = \pm 9

Step 1 - Isolate the quadratic term

In the given equation, the quadratic term is already isolated. The equation is: \[ x^2 = 81 \]

Step 2 - Apply the square root property

To solve for \( x \), take the square root of both sides of the equation: \[ \sqrt{x^2} = \sqrt{81} \]

Step 3 - Simplify the radicals

Applying the square root property, remember that \( \sqrt{x^2} = x \) and \( \sqrt{81} = 9 \). So, taking the square root of both sides, we get: \[ x = \pm 9 \] This means there are two solutions: \( x = 9 \) and \( x = -9 \).

Key Concepts

solving quadratic equations

Quadratic equations are polynomial equations of degree 2, generally written in the form \( ax^2 + bx + c = 0 \). To solve these, you can use various methods such as factoring, completing the square, using the quadratic formula, and for some specific forms, the square root property.

In this exercise, the equation is \(x^2 = 81 \), which is already a simplified form that suits the square root property method.

simplifying radicals

Simplifying radicals involves reducing a radical expression to its simplest form. In this exercise, we find the square root of both sides of the equation \(x^2 = 81 \).

Recall that the square root function has both a principal (positive) and a negative root. Thus, when taking the square root of 81, you need to account for both \(\text{+}9\) and \(\text{-}9\). Hence, this gives us two possible solutions: \(x = \text{+}9 \) and \(x = \text{-}9 \).

It's always important to remember this dual nature of square roots when solving quadratic equations.

isolating quadratic terms

Isolating the quadratic term means rearranging the equation so that the term with the square (\(x^2\)) is by itself on one side of the equation. For the given exercise, \(x^2 = 81 \), the quadratic term \(x^2\) is already isolated.

This step is crucial in making the equation simpler to solve, especially when using methods like the square root property. If the quadratic term wasn't isolated initially, you would need to perform algebraic operations to move other terms to the opposite side of the equation before proceeding with solving the equation.