Find two consecutive .....?

Question

Find two consecutive positive integers such that the sum of
their squares is 85.

Answer 1

first you need to find your variables.
the first number (the smaller one) will be X.
the second number is one more than X, which is (X+1).
when these numbers are squared, they equal 85, so
X^2+(X+1)^2=85
now write out the binomial square (X+1)(X+1)= X^2+2X+1
now your problem should read:
X^2+X^2+2X+1=85
combine the X^2 (the like terms)
2X^2+2X+1=85
now set one side equal to zero by subtracting the 85 from both sides.
2X^2+2X-84=0
now factor the trinomial
2(X+7)(X-6)
your solutions are -7 and 6
you don't want the negative answer, so your answer is
6!!!!!!!!!

Answer 2

6 and 7 are the numbers.
If you need to work it out, rather than guess and check, here's how:
The two consecutive numbers are n and n+1
n^2 + (n+1)^2 = 85
n^2 + n^2 + 2n + 1 = 85
2n^2 + 2n - 84 = 0
n^2 + n - 42 = 0
(n+7)(n-6) = 0
n = -7 or 6
You're looking for positive integers, so the answer is n= 6 and n+1=7.

The equation lets you know that -7, -6 are also two consecutive integers (just not positive ones ) whose squares add to 85.

Answer 3

Suppose the lower number is x, so the higher number is x+2, therefore: x(x + 2) = 143 x^2 + 2x - 143 = 0 x = (-2 +/- sqrt(2^2 - 4(1)(-143))) / (2*1) x = (-2 +/- sqrt(4 + 572)) / 2 x = (-2 +/- sqrt(576)) / 2 x = (-2 +/- 24) / 2 x = (22 or -26) / 2 x = 11 or -13 So the numbers could be 11 and 13, or they could be -13 and -11.

Answer 4

6 and 7(36 plus 49=85)