Find three consecutive even numbers, the sum of whose squares is 116.
Let the three consecutive even numbers be (x-2) , (x) , (x+2)
Acc to ques :
=> $$(x-2)^2 + x^2 + (x+2)^2 = 116$$
=> $$(x^2 - 4x + 4) + x^2 + (x^2 + 4x + 4) = 116$$
=> $$3x^2 = 116-8$$
=> $$x^2 = 36$$
=> $$x = \pm6$$
Since, numbers are even, they are all positive
=> Numbers are (6-2) , (6) , (6+2) = 4,6,8
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