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

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Answer:

6 and 7 are the consecutive numbers.

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
n and
n+1 are consecutive integers.

If they are both positive, then clearly
n+1 is more than
n.

We are given the square of the smaller,
n, is added to seven times the larger,
n+1, is equal to 85.

This means:


n^2+7(n+1)=85

Let's solve.

First we should use the distributive property:


n^2+7n+7=85

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
n.

Subtract 85 on both sides:


n^2+7n+7-85=0

Simplify:


n^2+7n-78=0

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:


(n-6)(n+13)=0

This implies
n-6=0 or
n+13=0.

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

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

Now we are looking for positive integers, so
n=6.

This implies the value of
n+1=6+1=7.

The numbers are 6 and 7.

Let's check.


6^2=36 <-This is the square of the smaller.


6^2+7(7)=36+49=85 <-This is the square of the smaller added to seven times the larger. We got 85 so the check is good.

User Sergey Alaev
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