Question 1081746: Find three consecutive positive integers such that the square of the first increased by the last is 22.
Please show the work.I know the answer(4,5,6) but the other tutor didn't show how he got this answer.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! 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)
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