The sum of the squares of 2 consecutive numbers is less than or equal to 25 find the possible values of the smallest integer?

Answer 1

x and x+1 are the numbers

the squares are x^2 and (x^2+2x+1)

We need these to be less and or equal to 25 so we write

2x^2+2x+1< or = to 25

2x^2+2x-24< or equal to 0

Solve 2(x^2+x-12)=0

2(x+4)(x-3)=0

so x=-4 or x=3

Now note that 2x^2+2x-24 is the graph of a parabola opening toward the positive y direction, that is to say, opening upward.

Now thinking of f(x)=2x^2+2x-24 we want the x values such that y is < and = to 0. In fact we want the smallest values.

If x is greater than -4 than y is less than 0, if x is less than 3, y is greater than 0.

So the answer is for the inequality in interval notation is [-4,3]. If we want the smallest integer that will work, it will be -4.

Let's try it

x=-4 and the next consecutive integer is -3.

16+9=25 so that works!

If x is 3 than the integers are 3 and 4 and once again it is 25. For any numbers between -4 and 3, say 0, it clearly works also, but if x is -5 and the other number is -4 then the sum is 25+16 which is greater.

So the smallest integers are -4 and -3.