How do you solve #15x^(-2)+7x^(-1)-2=0#?

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Answer 1 · 2015-11-06T06:43:09

Multiply through by #x^2# to turn it into a standard quadratic equation, and then solve with the quadratic formula

First, we multiply each side of the equation by #x^2#.

#x^2(15x^-2 + 7x^-1 - 2) = 0x^2#
#=>15 + 7x - 2x^2 = 0# (by applying the rule #x^a * x^b = x^(a+b)#)

Rearranging the terms gives us a familiar quadratic form.

#-2x^2 + 7x + 15 = 0#

Finally, we apply the quadratic formula #x = (-b +- sqrt(b^2 -4ac))/(2a)# where #a = -2#, #b = 7#, and #c = 15#

This gives

#x = -3/2# or #x = 5#.