How do you find two consecutive even integers such that the square of the smaller is 10 more than the larger?

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How do you find two consecutive even integers such that the square of the smaller is 10 more than the larger?

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Answer 1 · 2016-06-21T02:13:00

I found: #4 and 6#

Explanation:

Call the two numbers:
#2n#
and
#2n+2#
so we get:
#(2n)^2=(2n+2)+10#
#4n^2=2n+2+10#
#4n^2-2n-12=0#
solve for #n#:
#n_(1,2)=(2+-sqrt(4+192))/8=#
#n_(1,2)=(2+-14)/8#
#n_1=2#
#n_2=-12/8=-3/2#
so we have using #n_1#:

#4 and 6#

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How do you find two consecutive even integers such that the square of the smaller is 10 more than the larger?

1 Answer

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