Problem 90
Question
Solve the equation. $$x^{2}+25=81$$
Step-by-Step Solution
Verified Answer
The solutions to the equation \(x^{2}+25=81\) are \(x = 2\sqrt{14}\) and \(x = -2\sqrt{14}\).
1Step 1: Isolate the x Square Term
First, isolate the \(x^2\) on one side of the equation. To do this, subtract 25 from both sides of the equation to get: \(x^{2} + 25 - 25 = 81 - 25\), which simplifies to \(x^{2} = 56\).
2Step 2: Square Root both Sides
Next, find the value for \(x\), perform a square root on both sides of the equation as \(x\) is squared in the equation. However, remember when you apply square root on an equation, you get both positive and negative solutions. Therefore, \(x = \pm \sqrt{56}\).
3Step 3: Simplify the Square Root
The square root of 56 does not give a whole number. We can simplify it to \(x = \pm 2\sqrt{14}\).
Key Concepts
Isolating TermsSquare Root PropertySimplifying Radicals
Isolating Terms
Before diving into solving quadratic equations, let's focus on the concept of isolating terms. Isolation of terms is a crucial step when solving an equation as it sets the stage for more advanced operations.
The goal of isolating terms is to have the unknown variable on one side of the equation by itself. In our exercise, the equation is:
The goal of isolating terms is to have the unknown variable on one side of the equation by itself. In our exercise, the equation is:
- \( x^2 + 25 = 81 \)
- \( x^2 + 25 - 25 = 81 - 25 \)
- \( x^2 = 56 \)
Square Root Property
The square root property is used to solve equations where the variable is squared, like our example equation \( x^2 = 56 \). Utilizing this property involves taking the square root of both sides of the equation.
When we do this, it's vital to remember that taking the square root produces two potential outcomes: both positive and negative roots. This is a key point where students often make mistakes, so always keep in mind that:
Thus, the square root property remains a powerful tool for working through quadratic equations and finding complete solutions.
When we do this, it's vital to remember that taking the square root produces two potential outcomes: both positive and negative roots. This is a key point where students often make mistakes, so always keep in mind that:
- \( x = \pm \sqrt{56} \)
Thus, the square root property remains a powerful tool for working through quadratic equations and finding complete solutions.
Simplifying Radicals
After applying the square root property, you may find that the square root isn't a whole number. In such instances, simplifying radicals is beneficial for making equations cleaner and easier to handle in different mathematical contexts.
In our problem, we are left with \( x = \pm \sqrt{56} \). Since 56 isn't a perfect square, the square root doesn't simplify automatically to a simple number. Thus, we simplify it using factorization:
In our problem, we are left with \( x = \pm \sqrt{56} \). Since 56 isn't a perfect square, the square root doesn't simplify automatically to a simple number. Thus, we simplify it using factorization:
- Find the prime factors of 56: \( 56 = 2 \times 2 \times 14 \)
- Recognize that \( 2 \times 2 = 4 \) and \( \sqrt{4} = 2 \)
- Thus, \( \pm \sqrt{56} = \pm 2\sqrt{14} \)