How do you find three consecutive positive integers such that the square of the first, increased by the last, is 22?

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Answer 1 · 2015-10-19T10:17:37

Let's cal the integers #x,x+1,x+2#

Then #x^2+(x+2)=22#. Now subtract #22# on both sides:

#x^2+x-20=0-># which can be factorized into:

#(x+5)(x-4)=0->#

So either #(x+5)=0->x=-5# which is not positive,

Or #x-4=0->x=4# and this one is positive

Answer : #4,5,6#

Check : #4^2+6=22#