Equivalent Equation To √x²+81 = X+10 A Step-by-Step Guide
In the realm of mathematics, solving equations often involves transforming them into equivalent forms that are easier to manipulate and solve. This article delves into the process of identifying the equation equivalent to the given equation, √x²+81 = x+10. We will explore the steps involved in transforming the original equation and analyze the provided options to determine the correct equivalent equation.
Understanding the Original Equation
The equation we are given is √x²+81 = x+10. This equation involves a square root, which can make it slightly more challenging to work with directly. To simplify the equation, our primary strategy will be to eliminate the square root. This is typically achieved by squaring both sides of the equation. Squaring both sides ensures that we maintain the equality, as long as we consider the potential introduction of extraneous solutions, which we'll discuss later.
The left-hand side of the equation is √x²+81. When we square this, the square root is canceled out, leaving us with x²+81. The right-hand side of the equation is x+10. Squaring this requires us to expand the binomial, which we'll do using the distributive property or the FOIL method. This process is crucial for finding the correct equivalent equation among the provided options.
It's important to remember that squaring both sides of an equation can sometimes introduce solutions that don't satisfy the original equation. These are known as extraneous solutions. Therefore, after finding potential solutions to the squared equation, we must check them in the original equation to ensure they are valid. This step is critical for arriving at the correct solution set and avoiding errors in mathematical problem-solving.
Step-by-Step Transformation of the Equation
To find the equivalent equation, we'll square both sides of the original equation, √x²+81 = x+10. This is a fundamental algebraic technique used to eliminate square roots and simplify equations. The process involves applying the squaring operation to both the left-hand side (LHS) and the right-hand side (RHS) of the equation, ensuring that the equality is maintained.
Squaring the Left-Hand Side
The LHS of the equation is √x²+81. When we square this, the square root is effectively canceled out. Mathematically, this can be represented as follows:
(√x²+81)² = x²+81
The square root and the square operation are inverse operations, meaning they undo each other. This simplification is a key step in transforming the equation into a more manageable form. By eliminating the square root, we move closer to an equation that we can solve using standard algebraic techniques.
Squaring the Right-Hand Side
The RHS of the equation is x+10. Squaring this involves multiplying the binomial by itself, which can be done using the distributive property (also known as the FOIL method). The calculation is as follows:
(x+10)² = (x+10)(x+10)
Applying the distributive property, we get:
= x(x+10) + 10(x+10)
= x² + 10x + 10x + 100
Combining like terms, we obtain:
= x² + 20x + 100
This expansion is crucial for identifying the correct equivalent equation. The resulting quadratic expression will be compared to the options provided to determine which one matches our transformed equation.
Identifying the Equivalent Equation from the Options
After squaring both sides of the original equation √x²+81 = x+10, we arrived at the equation x²+81 = x²+20x+100. Now, we need to compare this transformed equation with the given options to identify the equivalent one. This process involves carefully examining each option and matching it against our result.
Analyzing the Options
Let's consider the options provided:
A. x+9 = x+10
This option is a linear equation and does not match the quadratic nature of our transformed equation. It is a simple equation that suggests a constant difference, which is not present in our derived equation. Therefore, option A is not the equivalent equation.
B. x+9 = x²+20x+100
This option presents a mix of linear and quadratic terms. While the right-hand side matches the quadratic expression we obtained, the left-hand side does not correspond to the left-hand side of our transformed equation. Thus, option B is not the equivalent equation.
C. x²+81 = x²+100
This option has quadratic terms on both sides, but the right-hand side is missing the 20x term that we found when squaring (x+10). This discrepancy indicates that option C is not the correct equivalent equation.
D. x²+81 = x²+20x+100
This option perfectly matches the equation we derived by squaring both sides of the original equation. The left-hand side (x²+81) and the right-hand side (x²+20x+100) are identical to our result. Therefore, option D is the equivalent equation.
Conclusion
Based on our step-by-step transformation and analysis, option D, x²+81 = x²+20x+100, is the equivalent equation to the original equation √x²+81 = x+10. This conclusion is reached by squaring both sides of the original equation and comparing the result with the provided options. This process highlights the importance of careful algebraic manipulation and comparison in solving mathematical problems.
Importance of Checking for Extraneous Solutions
While we've identified the equivalent equation, it's crucial to remember the importance of checking for extraneous solutions. Squaring both sides of an equation, especially one involving square roots, can introduce solutions that do not satisfy the original equation. These are known as extraneous solutions, and they arise because the squaring operation can mask the sign of the expressions involved.
Understanding Extraneous Solutions
Extraneous solutions occur because the squaring operation makes both positive and negative values of an expression equal. For example, both 3 and -3, when squared, result in 9. This means that when we square both sides of an equation, we might be introducing solutions that would make the original equation untrue.
In the context of our equation, √x²+81 = x+10, squaring both sides led us to x²+81 = x²+20x+100. While this new equation is equivalent in an algebraic sense, it may have solutions that do not work when plugged back into the original equation because the square root function only returns non-negative values.
The Process of Checking Solutions
To check for extraneous solutions, we need to solve the equivalent equation x²+81 = x²+20x+100 and then substitute each solution back into the original equation, √x²+81 = x+10. If a solution makes the original equation true, it is a valid solution. If it makes the original equation false, it is an extraneous solution and must be discarded.
Solving the Equivalent Equation
First, let's solve x²+81 = x²+20x+100. We can simplify this equation by subtracting x² from both sides:
81 = 20x + 100
Next, subtract 100 from both sides:
-19 = 20x
Finally, divide by 20:
x = -19/20
So, we have one potential solution: x = -19/20. Now, we must check if this solution is valid by substituting it back into the original equation.
Substituting the Solution
Substitute x = -19/20 into √x²+81 = x+10:
√((-19/20)² + 81) = -19/20 + 10
√((361/400) + 81) = -19/20 + 200/20
√((361/400) + (32400/400)) = 181/20
√(32761/400) = 181/20
181/20 = 181/20
Since the equation holds true, x = -19/20 is a valid solution and not an extraneous one. This step underscores the importance of verifying solutions when dealing with equations involving radicals.
Conclusion
In conclusion, the equation equivalent to √x²+81 = x+10 is x²+81 = x²+20x+100, which corresponds to option D. We arrived at this conclusion by squaring both sides of the original equation and comparing the resulting equation with the provided options. Furthermore, we emphasized the crucial step of checking for extraneous solutions to ensure the validity of our results. This comprehensive approach is essential for accurate mathematical problem-solving.
