Question 1081746
Let x be the first integer. The value of x is positive, so x > 0.


The next integer after x is x+1 which is the second number of the consecutive integer string. 


The last integer of this grouping is (x+1)+1 = x+1+1 = x+2


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The three integers, in terms of x, are
x
x+1
x+2


Keep in mind that x is just a placeholder for some unknown number. 


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We're told that "the square of the first increased by the last is 22", meaning that...


(first integer of group)^2 + (last integer of group) = 22


Using the variable expressions set up above, we would end up with


(x)^2 + (x+2) = 22


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Let's get everything to one side, simplify and then factor to solve for x


(x)^2 + (x+2) = 22
x^2 + x+2 - 22 = 0
x^2 + x - 20 = 0
(x+5)(x-4) = 0 ... see note below
x+5 = 0 or x-4 = 0
x = -5 or x = 4


Note: to factor, you need to find two numbers that multiply to -20 and add to 1. Those two numbers are +5 and -4. 


The two solutions are x = -5 or x = 4. However, recall that these integers are positive. Since x > 0, this rules out x = -5. The only practical solution is x = 4.


If x = 4, then
first integer = x = 4
second integer = x+1 = 4+1 = 5
third integer = x+2 = 4+2 = 6


Which is why the ordered triple solution is (4,5,6)