Find two consecutive even integers such that the square of the smaller is 10 more than the larger.
1 answer:
Let the two consecutive even integers be x and x+2.
Then x^2 = (x+2) + 10
Simplifying, x^2-x -2 - 10 = 0, or x^2 - x - 12 = 0. Factoring,
(x-4)(x+3) = 0. Solving for x: x=4 and x= -3.
If the first number is x = 4, then the second is x+2=4+2=6.
But wait! If the first number is -3, then the second is x+2 = -3+2 = -1.
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