Question

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

Answer 1

Answer:

The numbers are 5, 6, 7 and the equation used in the process of solving this problem is $3n^{2}+6n+5=110$

Step-by-step explanation:

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

You can use n to represent the first number because the statement says that the numbers are consecutive, we can say that the second number is represented by n+1 and the third by n+2.

Next, we need the squares of the numbers so, $n^{2} , (n+1)^{2}, (n+2)^{2}$.

And the last step is the sum of the squares to be equal to 110 $n^{2}+(n+1)^{2}+(n+2)^{2}=110$.

To find the equation, we expand and simplify the expression $n^{2}+(n+1)^{2}+(n+2)^{2}=110\\ n^{2}+n^{2}+2n+1+n^{2}+4n+4=110\\ 3n^{2}+6n+5=110$

So the equation used in the process of solving this problem is $3n^{2}+6n+5=110$

Part b: What are the numbers?

We solve the equation $3n^{2}+6n+5=110$ using the quadratic formula.

$3n^{2}+6n-105=0\\ x_{1,2} =\frac{-b\±\sqrt{b^{2}-4ac} }{2a}\\ x_{1}= \frac{-6+\sqrt{6^{2}-4*3*-105} }{2*3}, x_{2}= \frac{-6-\sqrt{6^{2}-4*3*-105} }{2*3}\\ x_{1}=5, x_{2}=-7$

We cannot use $x_{2}=-7$ because the problem says positive integers.

So the numbers are n=5, n+1=6, n+2=7

Answer 2

$n^2 + (n+1)^2 + (n+2)^2 = 110$

$n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 110$

$3n^2 + 6n + 5 = 100$

Answer: Choice C