The sum of the squares of 3 consecutive positive integers is 110. what are the numbers? which of the following equations is used in the process of solving this problem? a.3n 2 + 5 = 110 b.3n 2 + 3n + 3 = 110 c.3n 2 + 6n + 5 = 110

Part (a):

What are the numbers?

Let the numbers are ⇒ x , (x+1) and (x+2)

And the square of the numbers are ⇒ x² , (x+1)² and (x+2)²

∴ x² + (x+1)² + (x+2)² = 110

∴ x² + (x² + 2x + 1) + (x² + 4x + 4) = 110

∴ 3 x² + 6x + 5 = 110

∴ 3 x² + 6x - 105 = 0

divide the all terms over 3

∴ x² + 2x - 35 = 0 ⇒⇒ factor the equation

∴ (x + 7)(x - 5) = 0

∴ x = -7 ⇒(unacceptable because the numbers are positive)

OR x = 5

so, the numbers are 5 , 6 and 7

Part (b) :

Which of the following equations is used in the process of solving this problem?

Let the numbers are ⇒ x , (x+1) and (x+2)

And the square of the numbers are ⇒ x² , (x+1)² and (x+2)²

∴ x² + (x+1)² + (x+2)² = 110

∴ x² + (x² + 2x + 1) + (x² + 4x + 4) = 110

∴ 3 x² + 6x + 5 = 110

So, the correct answer is option c. 3n² + 6n + 5 = 110

damm bro the struggle

step-by-step explanation: