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put this solution on YOUR website!10x^-2 -3x^-1 -1=0
10/(x^2)-3/(x) -1 = 0 ----(1) (using a^(-m) = 1/(a^m) )
Multiplying by x^2 through out
10-3x-x^2 =0
Multiplying by -1 through out
-10+3x+x^2 = 0
That is x^2+3x-10 = 0 ----(2)
(rewriting the terms along with their correct signs
of course in the order we want)
The product ii (-10) and the sum is +3
Therefore the quantities are (+5) and (-2)
Therefore (2) is written as
x^2+5x-2x-10 = 0
(x^2+5x)-2x-10 = 0
x(x+5)-2(x+5) =0
xp-2p = 0 where p = (x+5)
p(x-2) = 0
That is (x+5)(x-2) = 0
(x+5) = 0 gives x= -5
(x-2) = 0 gives x= 2
Answer: x =-5 and x = 2
Verification: x = -5 in 10/(x^2)-3/(x) -1 = 0 ----(1)
LHS = 10/(x^2)-3/(x) -1
= 10/25 -3/(-5) - 1
=2/5+3/5-1 = (2+3)/5 -1 = 1-1 =0 =RHS
x = 2 in 10/(x^2)-3/(x) -1 = 0 ----(1)
LHS = 10/(x^2)-3/(x) -1
= 10/4 -3/2 - 1
=5/2-3/2-1 = (5-3)/2 -1 = 1-1 =0 =RHS
Therefore both our value are correct
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