The positive difference between two consecutive perfect squares is 59. What is the greater of the two perfect squares?

APatel Feb 8, 2019

#1**+1 **

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. |

TheXSquaredFactor Feb 8, 2019

#3**+1 **

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

ElectricPavlov Feb 8, 2019