Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 214. Which are those three integers?

Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 214. Which are those three integers?

1. A
6,7,8
2. B
4,5,6
3. C
7,8,9
4. D
5,6,7

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Solution:

Let the three consecutive positive integers be x, x+1 and x+2.
Given, ${x}^{2}$+${\left(x+1\right)}^{2}$+${\left(x+2\right)}^{2}$= ${\left(x+x+1+x+2\right)}^{2}$−214
${x}^{2}$+${x}^{2}$+1+2x+${x}^{2}$+4+4x=${\left(3x+3\right)}^{2}$−214
⇒3${x}^{2}$+6x+5=9${x}^{2}$+9+18x−214
⇒3${x}^{2}$+6x+5=9${x}^{2}$+18x−205
⇒6${x}^{2}$+12x−210=0
${x}^{2}$+2x−35=0
${x}^{2}$+7x−5x−35=0
⇒x(x+7)−5(x+7)=0
⇒(x+7)(x−5)=0
Then, x+7=0 or x−5=0
⇒x=−7 or x=5
Given that the numbers are positive.
Hence, x cannot be −7
Then, x=5
So, x+1=5+1=6
and x+2=5+2=7
then, The three consecutive positive integers are 5, 6 and 7. Hence (4) is the correct option.

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