Question

Find two consecutive positive intergers such that the square of the smaller interger is nineteen more than five times the larger interger.

Answer 1

Answer:

8 and 9

Step-by-step explanation:

Let the two consecutive integers be x - 1 and x

Smaller = x-1

Largere = x

If the square of the smaller integer is nineteen more than five times the larger integer, then;

(x-1)² = 19 + 5x

Expand

x²-2x+1= 19+5x

x²-2x-5x+1-19 -0

x²-7x-18 =0

Factorize

x²-9x+2x-18 =0

x(x-9)+2(x-9) = 0

(x+2) = 0 and x - 9 = 0

x = -2 and 9

SInce x cannot be negative;

x = 9

Smaller number = 9-1

Smaller number = 8

Hence the two consecutive integers are 8 and 9