Find 3 consecutive integers such that twice the greatest integer is 2 less than 3 times the least integer

The consecutive integers are n, n+1, n+2

2(n+2)=3n-2 perform indicated multiplication on left side

2n+4=3n-2 subtract 3n from both sides

-n+4=-2 subtract 4 from both sides

-n=-6 divide both sides by -1

n=6

So the three consecutive numbers are 6,7, and 8

the polynomial is a perfect square trinomial of the form a^2-2ab+b^2:

option a. true

step-by-step explanation:

9x^2-12x+4

comparing with the form a^2-2ab+b^2

1.) a^2=9x^2

solving for a: square root both sides of the equation:

sqrt(a^2)=sqrt(9x^2)

a=sqrt(9) sqrt(x^2)

a=3x

2.) b^2=4

solving for b: square root both sides of the equation:

sqrt(b^2)=sqrt(4)

b=2

to be a perfet square trinomial 2ab must be equal to 12x. let's check:

2ab = 2 (3x) (2)

2ab = 12x

correct