SOLUTION: Find the real or imaginary numbers of the equation: x^-2

SOLUTION: Find the real or imaginary numbers of the equation: x^-2 - 9x^-1 + 18=0

Question 97352: Find the real or imaginary numbers of the equation:
x^-2 - 9x^-1 + 18=0
Found 2 solutions by jim_thompson5910, stanbon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Flip all the variables with negative exponents

Multiply both sides by the LCD

Distribute and multiply

Rearrange the terms

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve ( notice , , and )

Plug in a=18, b=-9, and c=1

Negate -9 to get 9

Square -9 to get 81 (note: remember when you square -9, you must square the negative as well. This is because .)

Multiply to get

Combine like terms in the radicand (everything under the square root)

Simplify the square root (note: If you need help with simplifying the square root, check out this solver)

Multiply 2 and 18 to get 36

So now the expression breaks down into two parts

or

Lets look at the first part:

Add the terms in the numerator
Divide

So one answer is

Now lets look at the second part:

Subtract the terms in the numerator
Divide

So another answer is

So our solutions are:
or

Notice when we graph , we get:

and we can see that the roots are and . This verifies our answer
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
x^-2 - 9x^-1 + 18=0
(x^-1-3)(x^-1-6)=0
x^-1= 3 or x^-1=6
x=1/3 or x=1/6
=================
Ceers,
Stan H.
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