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

Question

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

Answer 1

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)²  = 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

Comments

Thank you!!!