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Solve 1/x^3 + 8/x^2 + 16/x =0?

+1 vote
just started algebra and this section is about quadratic functions and other equations. this question is on a test and the book does not go over the answer if any can help me solve this i would appreciate it. Or at least point out the rules or how to find out how to solve this correctly.

the division signs mean its fractions.1 over x to the third power +8 over x to the second power plus 16 over x =0
asked Jan 21, 2013 in ALGEBRA 2 by dkinz Apprentice

1 Answer

+1 vote
 
Best answer

1/x3+8/x2+16/x=0

Multiply each side by x3.

x3(1/x3+8/x2+16/x) = x3(0)

Simplify

x3(1/x3) + x3(8/x2) + x3(16/x) = 0

1 + 8x + 16x2 = 0

16x2 + 8x + 1 = 0

This equation it can be written as

(4x)2 + 2(4x)(1) + (1)= 0

Compare the equation a2 + 2ab + b2 = (a+b)2

So ,

(4x + 1)2 = 0

(4x + 1)(4x + 1) = 0

4x + 1 = 0  or 4x + 1 = 0

4x + 1 = 0

Subtract 1 from each side.

4x + 1 - 1 = -1

4x = -1

Divide each side by 4

4x / 4 = -1 / 4

x = - 1/ 4

There fore x = - 1/ 4

answered Jan 22, 2013 by richardson Scholar
selected Jan 23, 2013 by dkinz
Thank you!!!

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