Two consecutive positive numbers are such that the sum of their squares is 113. find the two numbers

7 and 8

Step-by-step explanation:

Let x represent the first number, then x + 1 is the other number.

(x)² + (x + 1)² = 113

x²  +  x² + 2x + 1 = 113    expanded (x + 1)²

2x² + 2x + 1 = 113          added like terms

2x² + 2x - 112 = 0        subtracted 113 from both sides

x² + x - 56 = 0           divided both sides by 2

(x + 8) (x - 7) = 0        factored polynomial

x + 8 = 0   x - 7 = 0    applied zero product property

x = -8     x = 7            solved for x

  ↓

not valid since the restriction is that x > 0 (a positive number)

So, x = 7     and     x + 1 = (7) + 1   = 8

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)