Question 193

Find three consecutive even numbers, the sum of whose squares is 116.

Solution

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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