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

 Sep 14, 2018
 #1
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+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

 Sep 15, 2018
 #2
avatar+27775 
+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

 Sep 15, 2018

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