Question 1204833
Find two positive even consecutive integers such that one less than the square of the smaller integer is three more than the four times the larger integer.

Let the two integersbe x, x+2

one less than the square of the smaller integer is three more than the four times the larger integer.

x^2-1 = 4(x+2)+3

x^2-1= 4x+8+3

x^2-1= 4x+11

x^2-4x-11-1=0

x^2-4x-12=0

factorise

x^2-6x+2x-12=0

x(x-6)+2(x-6)=0

(x-6)(x+2)=0

x =6  or x=-2

x cannot be -2 because the intergers are positive (given)

So the integers are 6, 8

Check

6^2-1 =35= 4(8)+3