Question

sum of a no. and its reciprocal is 3 41/80 find the no.

Answer 1

Let the number be x.

Therefore the reciprocal = 1/x

Therefore x + 1/x = 3 41/80 = 281/80 

x + 1/x =   x² + 1     = 281/80 
                  x    
      
Cross multiply   281x = 80x² + 80
Therefore:

80x² - 281x + 80 = 0

Expand the equation

80x² - 256x - 25x + 80 = 0
16x(5x - 16) - 5(5x - 16) = 0
(16x - 5) (5x - 16) = 0

Therefore:    

16x - 5 = 0      or      5x - 16 = 0 
16x = 5   or 5x = 16
16x = 5 , hence x = 5/16
    or 
5x = 16,  hence x = 3 1/5 (3 and 1/5)
Therefore x = 5/16  or 16/5

Answer 2

Let the number be x and 1/x
Therefore, according to the question,
x + 1/x = 3 41/80
x² + 1 
______  = 281/80
    x
80x² + 80 = 281x
80x² - 281x + 80 = 0
x = 281 ⁺  (√281² - 4×80×80)/2×80
             ⁻
= 281 ⁺ (√78961 - 25600)/160
          ⁻ 
= 281 ⁺ (√53361)/160
          ⁻ 
= (281 ⁺ 231)/160
           ⁻
= (281+231)/160 and (281 - 231)/160

= 512/160 and 50/160

16/5 and 5/16

The two numbers are 16/5 and 5/16
Answer