+0  
 
0
21
2
avatar+149 

Suppose the decimal \(0.28\) is equal to \(\frac{1}{n}+\frac{2}{n^2}\), where \(n\) is negative. Find \(n\).

 
Logic  Sep 14, 2018
 #1
avatar
+1

Solve for n:
1/n + 2/n^2 = 0.28

0.28 = 7/25:
1/n + 2/n^2 = 7/25

Bring 1/n + 2/n^2 together using the common denominator n^2:
(n + 2)/n^2 = 7/25

Cross multiply:
25 (n + 2) = 7 n^2

Expand out terms of the left hand side:
25 n + 50 = 7 n^2

Subtract 7 n^2 from both sides:
-7 n^2 + 25 n + 50 = 0

The left hand side factors into a product with three terms:
-(n - 5) (7 n + 10) = 0

Multiply both sides by -1:
(n - 5) (7 n + 10) = 0

Split into two equations:
n - 5 = 0 or 7 n + 10 = 0

Add 5 to both sides:
n = 5 or 7 n + 10 = 0

Subtract 10 from both sides:
n = 5 or 7 n = -10

Divide both sides by 7:
 
n = 5     or       n = -10/7

 
Guest Sep 15, 2018
 #2
avatar+26965 
+2

1/n + 2/n2 = 0.28

 

Let x = 1/n

 

Then 2x2 + x - 0.28 = 0  which factors as (2x + 1.4)(x - 0.2)

 

The negative solution is obtained from 2x + 1.4 = 0 or x = - 0.7

 

hence n = 1/x = -1/0.7  or  n = -10/7 or n ≈ -1.429

 
Alan  Sep 15, 2018

24 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.