Find two consecutive positive integers such that the square of the smaller integer added to seven times the larger integer is equal to 85

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 and are consecutive integers.

If they are both positive, then clearly is more than .

We are given the square of the smaller, , is added to seven times the larger, , is equal to 85.

This means:

Let's solve.

First we should use the distributive property:

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 .

Subtract 85 on both sides:

Simplify:

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:

This implies or .

We are going to solve the first equation by adding 6 on both sides giving us: .

We are going to solve the second equation by subtracting 13 on both sides giving us: .

Now we are looking for positive integers, so .

This implies the value of .

The numbers are 6 and 7.

Let's check.

<-This is the square of the smaller.

<-This is the square of the smaller added to seven times the larger. We got 85 so the check is good.

a is the right answer

answer: i think you should do the math and pay attention in class : )

step-by-step explanation:

that is truly what   think but if i get the answer i will