A number added to another number and the result divided by a square is the same as a square times the number. Find the number.
Answer (1)
Let the number be x . The problem translates to the equation y 2 x = y x . Solving for x , we get x ( y 2 − 2 ) = 0 . Thus, either x = 0 or y 2 = 2 . The simplest solution is x = 0 . Therefore, the number is 0 .
Explanation
Setting up the equation Let's denote the unknown number as x . According to the problem, we add the number to itself, which gives us x + x = 2 x . Then, we divide this result by some value, let's call it y . So we have y 2 x . The problem states that this is equal to y times the original number, which is y x . Therefore, we can write the equation as y 2 x = y x .
Solving for x Now, let's solve the equation y 2 x = y x for x . First, we multiply both sides of the equation by y to get rid of the fraction: 2 x = y 2 x . Next, we want to isolate x , so we rearrange the equation to get all terms involving x on one side: y 2 x − 2 x = 0 . Now, we can factor out x from the left side: x ( y 2 − 2 ) = 0 .
Finding possible solutions From the equation x ( y 2 − 2 ) = 0 , we have two possibilities: either x = 0 or y 2 − 2 = 0 . If x = 0 , then the equation holds true regardless of the value of y (as long as y = 0 , since we cannot divide by zero in the original equation). If y 2 − 2 = 0 , then y 2 = 2 , which means y = ± 2 . In this case, x can be any number.
Determining the number Therefore, the number x can be 0. If x is not 0, then y must be 2 or − 2 . However, since the problem does not specify any restrictions on y , the simplest solution is x = 0 .
Examples
Consider a scenario where you are trying to balance an equation in chemistry or physics. If you have a certain quantity of a substance and you manipulate it in a way that the final amount is proportional to the initial amount, this problem helps you understand the possible initial quantities that would satisfy the equation. For example, if you double the quantity of a reactant and then divide by a factor, and the result is that factor times the original quantity, finding the original quantity becomes an exercise in solving this type of equation.
