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Question 825367: find three consecutive positive integers such that the square of the first increased by twice the second is 3 less than four times the third.
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! ---
z = x + 2
y = x + 1
xx + 2y = 4z - 3
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xx + 2y = 4z - 3
xx + 2(x + 1) = 4(x + 2) - 3
xx + 2x + 2 = 4x + 8 - 3
xx - 2x - 3 = 0
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the above quadratic equation is in standard form, with a=1, b=-2, and c=-3
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
1 -2 -3
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
x = 3
x = -1
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the negative root doesn't fit the problem statement, so use the positive root
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answer:
the integers are: 3, 4, and 5
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