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# help

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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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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

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
+27775
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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