Word Problem: Find All Numbers

Question

find all numbers such that the sum of the number and half its reciprocal is 57/40

Answer 1

Let x be such a number...

 

so the problem tells us that x+1/2*1/x=57/40...that is to say 40(x+1/(2x))=57

or 57=40x+20/x...we need to note that x cannot be 0, or a reciprocal is hard to take...

 

Multiplying this whole equation by x yields:

57x=40x^2+20...this is a quadratic equation...and the solutions can be found with the quadratic formula!

 

0=40x^2-57x+20 has roots 

x=[57+/-sqrt(57^2-4*40*20)]/80

 

for the discriminant, b^2-4ac: 57^2-3200=49

 

so x=50/80 and 64/80...

 

lets check our work

 

for 50/80...half the reciprocal is 1/2*80/50=40/50...what is 50/80+40/50?

you can 64/80 in a similar fashion.

Answer 2

Let x = such a number

 

x + (1/2)(1/x) = 57/40

 

x + 1/(2x) = 57/40

 

Multiply the equation by the LCD, 40x, to get 40x² + 20 = 57x

 

40x² - 57x + 20 = 0

 

(8x - 5)(5x - 4) = 0

 

x = 5/8 or 4/5