How do you solve #(x-5)^2 + 3(x-5) + 9=0#?

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Answer 1 · 2017-11-09T14:22:36

#"The roots are "7/2+-3/2sqrt3i, or, (7+-3sqrt3i)/2.#

# (x-5)^2+3(x-5)+9=0.#

#:. (x-5)^2+2(x-5)(3/2)=-9.#

Completing the Square on the L.H.S., we get,

# (x-5)^2+2(x-5)(3/2)+(3/2)^2=-9+(3/2)^2=-27/4.#

#:. {(x-5)+3/2}^2=(-1)(3sqrt3/2)^2=(i^2)(3sqrt3/2)^2.#

Taking Square-Root of both sides,

# {(x-5)+3/2}=+-3/2sqrt3i.#

# :. x-7/2=+-3/2sqrt3i.#

#:. x=7/2+-3/2sqrt3i, or, x=(7+-3sqrt3i)/2,# are the desired roots.