Quadratic Formula for quadratic-type equations

[Kristen_Opp]

I am in finite math and we are reviewing using the quadratic formula. I'm stuck on two problems, it has been a long time since I've used the QF and am a little rusty.. help!??!!

1. X-4 - 9x-2 + 20 =0 (it is to the -4 power and same with the -2) I don't know how to set it up, or whether I should use the quadratic formula or factor it???

2. 2/x2 +3/x -2=0 (the denominator in the first term is X to the second power).
got rid of the fractions and reduced it to:
2x+3x2 (this is to the second power) -2=0 but I must have done something wrong because when I plug it into the QF I don't get what the back of the book says...

 

[skeeter]

x^-4 - 9x^-2 + 20 = 0

multiply all terms by x⁴ ...

1 - 9x² + 20x⁴ = 0

or ...

20⁴ - 9x² + 1 = 0

this equation, quadratic in x², will factor ...

(5x² - 1)(4x² - 1) = 0

now, set each factor equal to 0 and solve for x ... you will get 4 solutions.

------------------------------------------------------------------------------------

for the 2nd problem ... yes, you did something wrong.

2/x² + 3/x - 2 = 0

multiply all terms by x² ...

2 + 3x - 2x² = 0

this puppy will also factor ...

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

set each factor equal to 0 and solve for x ... yes, 2 solutions this time.


and you didn't even need to use the quadratic formula ... crisis averted.

 

[pka]

For #1. Let \(\displaystyle y = x^{-2}\) then you can solve \(\displaystyle y^2 -9 y + 20 = 0.\)
The go back and find x.