Respuesta :
Answer:
6 and 7 are the consecutive numbers.
Step-by-step explanation:
5 and 6 are consecutive integers because 6 is right after 5.
Let me revise the way I say that:
5 and 5+1 are consecutive integers.
So [tex]n[/tex] and [tex]n+1[/tex] are consecutive integers.
If they are both positive, then clearly [tex]n+1[/tex] is more than [tex]n[/tex].
We are given the square of the smaller, [tex]n[/tex], is added to seven times the larger, [tex]n+1[/tex], is equal to 85.
This means:
[tex]n^2+7(n+1)=85[/tex]
Let's solve.
First we should use the distributive property:
[tex]n^2+7n+7=85[/tex]
Now I would like the right hand side to be 0 so I can factor and then put both factors equal to 0 and solve for [tex]n[/tex].
Subtract 85 on both sides:
[tex]n^2+7n+7-85=0[/tex]
Simplify:
[tex]n^2+7n-78=0[/tex]
Now we want to find two numbers that multiply to be -78 and add to be 7.
Let's think of some numbers that multiply to be -78:
-78=-2(39); this wouldn't work because -2+39 isn't 7
-78=2(-39); this wouldn't work because 2+(-39) isn't 7.
-78=-6(13); this would work because -7+13 is 7.
So the factored form is:
[tex](n-6)(n+13)=0[/tex]
This implies [tex]n-6=0[/tex] or [tex]n+13=0[/tex].
We are going to solve the first equation by adding 6 on both sides giving us: [tex]n=6[/tex].
We are going to solve the second equation by subtracting 13 on both sides giving us: [tex]n=-13[/tex].
Now we are looking for positive integers, so [tex]n=6[/tex].
This implies the value of [tex]n+1=6+1=7[/tex].
The numbers are 6 and 7.
Let's check.
[tex]6^2=36[/tex] <-This is the square of the smaller.
[tex]6^2+7(7)=36+49=85[/tex] <-This is the square of the smaller added to seven times the larger. We got 85 so the check is good.
