Question

find the three consecutive integers positive integers such that sum of the square of first and product of the second and the third is 191​

Answer 1

Step-by-step explanation:

let the three consecutive positive integers be x-1,x,x+1

Acc to ques

(x-1)^2+x*x+1=191

x^2+1-2x+x^2+x-191=0

2x^2-x-190=0.

2x^2-20x+19x-190=0

2x(x-10)+19(x-10)=0

(2x+19)(x-10)=0

x=10 or x=-19/2

but since x is non negative therefore we will take x=10

x-1=10-1=9

x+1=10+1=11

The three consecutive positive integers are 9, 10and 11

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