[tex] n^2 + (n+1)^2 + (n+2)^2 = 110[/tex]
[tex] n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 = 110[/tex]
[tex] 3n^2 + 6n + 5 = 100 [/tex]
Answer: Choice C
Answer:
The numbers are 5, 6, 7 and the equation used in the process of solving this problem is [tex]3n^{2}+6n+5=110[/tex]
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, [tex]n^{2} , (n+1)^{2}, (n+2)^{2}[/tex].
And the last step is the sum of the squares to be equal to 110 [tex]n^{2}+(n+1)^{2}+(n+2)^{2}=110[/tex].
To find the equation, we expand and simplify the expression [tex]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[/tex]
So the equation used in the process of solving this problem is [tex]3n^{2}+6n+5=110[/tex]
Part b: What are the numbers?
We solve the equation [tex]3n^{2}+6n+5=110[/tex] using the quadratic formula.
[tex]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[/tex]
We cannot use [tex]x_{2}=-7[/tex] because the problem says positive integers.
So the numbers are n=5, n+1=6, n+2=7
