Question

Find two consecutive integers such that the sum of their squares added to the lesser number equals 21

Answer 1

Two consecutive numbers are x and x+1.

Clearly, x is the lesser, and x+1 is the largest.

The sum of their squares is

`x^2+(x+1)^2 = x^2+x^2+2x+1 = 2x^2+2x+1`

If we add this sum and the lesser we have

`2x^2+2x+1+x=2x^2+3x+1`

We want this quantity to be 21, so we have

`2x^2+3x+1=21 \iff 2x^2+3x-20=0`

The solutions of this equation are

`x=-4,\quad x=(5)/(2)`

Since we want integers, the numbers are -4 and -3.

Answer 2

-4 and -3.

Explanation:

Let the numbers be x and x+1.

From the given information:

x^2 + (x + 1)^2 + x = 21

x^2 + x^2 + 2x + 1 + x - 21 = 0

2x^2 + 3x - 20 = 0

(2x - 5 )(x + 4 ) = 0

x = 5/2 or -4 We ignore 5/2 as its not an interger

So the required integers are -4 and -3.