Question

Find a number such that if 25,7 and 1 are added to it,the product of the first and third results is equal to the square of the second.​

Answer 1

Answer:

Let the required number be x.

Given condition is

$(x + 25)(x + 1) = {(x + 7)}^{2} \\ \\ {x}^{2} + 26x + 25 = {x}^{2} + 14x + 49 \\ \\ {x}^{2} + 26x + 25 - {x}^{2} - 14x - 49 \\ = 0 \\ \\ 26 - 14x = 49 - 25 \\ \\ 12x = 24 \\ \\ \therefore \: x = \frac{ \cancel{24} \: \: ^{2} }{ \cancel{12}} \\ \\ \boxed{ \sf \blue{x = 2}}$

Hope it helps you from my side