consecutive integers?

Question

I have no idea how to do this, and it's a pretty long problem. I'd be eternally grateful if some kind soul out there can help me.

There are three consecutive positive integers such that the sum of the squares of the smallest two is 221.

Write an equation to find the three consecutive positive integers. (let x = the smallest integer)

Solve the equations, and write the three consecutive positive integers.

Answer 1

The consecutive integers would be x, x+1, and x+2.

x^2 + (x+1)^2 = 221 represents your statement about them

x^2 + x^2 + 2x + 1 = 221

2x^2 + 2x -220 =0

x^2 + x - 110=0

(x+11)(x-10) = 0

so x = 10 and x+ 1 = 11 and x + 2 = 12

Answer 2

The smallest integer is x, so the next two will be (x + 1) and (x + 2).

The sum of the squares of the smallest two is 221, so we have the following equation:

x^2 + (x + 1)^2 = 221.

Expanding gives

x^2 + (x^2 + 2x + 1) = 221.

Bringing everything to one side gives

2x^2 + 2x - 220 = 0.

Now let's factor:

2(x - 10)(x + 11) = 0.

This gives the two solutions x = 10, x = -11. However, since we know the numbers are all positive, then x = -11 doesn't work.

Therefore, the smallest integer must be 10, so our integers are 10, 11, and 12.

Answer 3

So, x^2 + (x+1)^2 = 221

Multiply. That gives you 2x^2 + 2x + 1 = 221

Subtract 221 from both sides and factor out 2.

You are left with x^2 + x -110 =0 OR (x +11) (x-10) =0.

X = -11 violates the positive integer rule.

So x = 10. Your three consecutive integers are 10, 11, 12.

Answer 4

10,11,12

If x, x+1, x+2 are the numbers, then:

x^2 + (x+1)^2 = 221
x^2 + x^2 + 2z + 1 = 221
2x^2 + 2x - 220 = 0
x^2 + x - 110 = 0
(x+11)(x-10) = 0
x = -11 or x = 10, and you wanted positive integers,
so x = 10