Find 2 consecutive positive integers such that the sum of their squares is 85.?

Answer 1

x² + (x + 1)² = 85
x² + x² + 2x + 1 = 85
2x² + 2x - 84 = 0
x² + x - 42 = 0
(x + 7)(x - 6) = 0
x = -7 or x = 6
x is not negative, so it is the 6
So x = 6 and x + 1 = 7
The numbers are 6 and 7

CHECK :
6² + 7² = 36 + 49 = 85

Answer 2

Let the smaller No. be x and then next number is (x+1)
x^2 + (x+1)^2 = 85
x^2 + x^2 + 2x +1 = 85
2x^2 + 2x – 84 = 85
x^2 + x – 42 = 0
x^2 – 6 x + 7– 42 = 0
x(x – 6) + 7(x – 6)= 0
(x – 6)(x + 7) = 0
(x – 6) = 0 or (x + 7) = 0
x = 6 or x = – 7
but number is +ve
so numbers are 6 and 7

Answer 3

y = first integer
x = 2nd integer

y^2 + x^2 = 85 --------eqn 1

x - y = 1; x = 1 +y -------eqn 2

y^2 + (1 +y)^2 = 85
y^2 + 1 + 2y + y^2 = 85
2y^2 +2y + 1 = 85
2(y^2 + y) = 84
y^2 + y - 42 = 0
( y + 7)(y - 6) = 0 ; since we need positive integers

we the positive root , y = 6
2nd integer, x = 7

Answer 4

x² + (x+1)² = 85
2x² + 2x + 1 = 85
2x²+2x = 84
x²+x -42 = 0
(x+7)(x-6)=0
So x = -7 or x = 6
Rejecting the negative root,
x = 6, x+1 = 7.
The answer is 6 and 7.

Answer 5

6 and 7

6^2=36
7^2=49

36+49=85

Answer 6

6, 7....pretty easy