What is the solution of [tex]$\sqrt{x+12}=x$[/tex]
A. [tex]$x=-3$[/tex]
B. [tex]$x=4$[/tex]
C. [tex]$x=-3$[/tex] or [tex]$x=4$[/tex]
D. no solution
Answer (2)
The solution to the equation x + 12 = x is x = 4 . The alternative solution x = − 3 does not satisfy the original equation. Thus, the correct answer is option B: x = 4 .
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Square both sides of the equation: ( x + 12 ) 2 = x 2 , which simplifies to x + 12 = x 2 .
Rearrange into a quadratic equation: x 2 − x − 12 = 0 .
Factor the quadratic equation: ( x − 4 ) ( x + 3 ) = 0 , giving possible solutions x = 4 and x = − 3 .
Check the solutions in the original equation: only x = 4 is a valid solution, since 4 + 12 = 4 .
The solution is 4 .
Explanation
Understanding the Problem We are given the equation x + 12 = x . Our goal is to find the solution(s) for x from the given options. We need to check each possible solution to see if it satisfies the original equation.
Eliminating the Square Root First, let's square both sides of the equation to eliminate the square root: ( x + 12 ) 2 = x 2 . This simplifies to x + 12 = x 2 .
Rearranging the Equation Now, rearrange the equation into a quadratic equation: x 2 − x − 12 = 0 .
Factoring the Quadratic Equation Next, factor the quadratic equation: ( x − 4 ) ( x + 3 ) = 0 .
Solving for x Solve for x : x = 4 or x = − 3 . These are the possible solutions.
Checking the Solutions Now, we need to check the solutions in the original equation x + 12 = x because squaring both sides can introduce extraneous solutions.
For x = 4 : 4 + 12 = 16 = 4 , which is true. So, x = 4 is a valid solution.
For x = − 3 : − 3 + 12 = 9 = 3 = − 3 , which is false. So, x = − 3 is not a valid solution.
Final Answer Therefore, the only solution that satisfies the original equation is x = 4 .
Examples
When solving problems involving distances or lengths, we often encounter square roots. For instance, if you're calculating the distance between two points and that distance is expressed as the solution to an equation involving a square root, you would need to solve the equation similarly to what we did here. Checking the solutions is crucial because distances cannot be negative, just like how we verified that x=4 was a valid solution while x=-3 was not.
