Question 958429: find three consecutive integers such that the square of the smallest increased by four times the largest is five.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! your smallest number is x
your next smallest number is x + 1
your largest number is x + 2
the square of the smallest increased by 4 times the largest is equal to 5 means:
x^2 + 4 * (x + 2) = 5
x is the smallest.
x + 2 is the largest.
simplify this equation to get:
x^2 + 4x + 8 = 5
subtract 5 from both sides of this equation to get:
x^2 + 4x + 3 = 0
factor this quadratic equation to get:
x = -3
x = 1
both of those values of x will satisfy the requirements of the problem.
when x = -3, x + 2 = -1.
x^2 + 4 * (x + 2) becomes (-3)^2 + 4 * (-1) which becomes 9 - 4 which becomes 5.
when x = -1, x + 2 = 1.
x^2 + 4 * (x + 2) becomes (-1)^2 + 4 * (1) which becomes 1 + 4 which becomes 5.
both x = -3 and x = -1 satisfy the requirements of the problems.
they are your solutions.
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