The positive difference between two consecutive perfect squares is 59. What is the greater of the two perfect squares?
Hello, APatel!
A perfect square is the product of a whole number multiplied by itself. Below is a list of the first few perfect squares, in order.
x | x^2 |
1 | 1 |
2 | 4 |
3 | 9 |
4 | 16 |
5 | 25 |
6 | 36 |
7 | 49 |
8 | 64 |
9 | 81 |
Let x equal a positive integer. Let x^2 equal any perfect square.
Let x+1 equal the subsequent positive integer. Let (x+1)^2 equal the next perfect square.
Notice how this is very similar to how I generated consecutive perfect squares in the table; I incremented the "x" column by 1 and squared the result. Also, notice that the question asks for the positive difference of the perfect squares. \((x+1)^2>x^2\) because (x+1)^2 represents the next perfect square. Therefore, it is possible to generate the following equation for this problem:
\((x+1)^2-x^2=59\) | Expand the binomial. |
\(x^2+2x+1-x^2=59\) | Look at that! The quadratic terms cancel out, and we are left with quite a basic equation. |
\(2x+1=59\) | Now, subtract one on both sides. |
\(2x=58\) | Divide by 2 on both sides. |
\(x=29\) | |
We are not done yet! Remember, our goal is to find the greater of the perfect squares. Of course, in this problem, I let (x+1)^2 represent the larger one.
\(x=29\\ (x+1)^2\) | Substitute 29 into this expression and simplify accordingly. |
\((29+1)^2\) | |
\(30^2\) | |
\(900\) | This is the larger of the 2 perfect squares. |
A lot of times, you can eliminate a step by assigning 'x' the value you are looking for....
x = larger number
x-1 = the previous number
The difference in these squares = 59
x^2 - (x-1)^2 = 59 expand
x^2 - x^2 + 2x -1 = 59
2x-1 = 59
2x = 60
x = larger number = 30 x^2 = 900