Question

The sum of the squares of two consecutive positive integers exceeds their product by 91

Answer 1

Answer:

Step-by-step explanation:

BRAINLIEST BRAINLIEST BRAINLIEST

Answer 2

Let the positive integers be x and x+1

According to Question ,

x^2 + (x+1)^2=x(x+ 1)+91

=>x^2 + x^2 + 2x + 1 = x^2 +x + 91

=> x^2 + x - 90 = 0

=> x^2+10x-9x-90 = 0

=>x (x+10)-9 (x+ 10) = 0

=>(x- 9)(x+ 10)= 0

So ,

x -9=0 and x+10=0

x= 9 and x = -10

Since we need positive integers , so

required integers are 9 and 9+1= 10