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2.

 Oct 31, 2018
 #1
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\(\dfrac 8 q + 2 = \dfrac{q+4}{q-1} \\ \dfrac{8+2q}{q} = \dfrac{q+4}{q-1} \\ (8+2q)(q-1) = q(q+4) \\ 2q^2+6q-8 = q^2 + 4q \\ q^2 + 2q-8=0 \\ (q+4)(q-2) = 0 \\ q= -4, ~2\)

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 Oct 31, 2018

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