How do you solve #x^2+8x-12=0#?

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Answer 1 · 2016-12-12T01:35:54

#x=-2-2sqrt7# or #x=-2+2sqrt7#

We can solve #x^2+8x-12=0# by using completing square method, as follows.

#x^2+8x-12=0# can be written as

#x^2+2×4×x+4^2-4^2-12=0#

or #(x+2)^2-16-12=0#

or #(x+2)^2-28=0#

or #(x+2)^2-(sqrt28)^2=0#

and as factors of #a^2-b^2# are #(a+b)# and #(a-b)#, quadratic polynomial on LHS can be factorized as under:

#(x+2+sqrt28)(x+2-sqrt28)=0#

and hence either #x+2+sqrt28=0# i.e. #x=-2-sqrt28=-2-2sqrt7#

or #x+2-sqrt28=0# i.e. #x=-2+sqrt28=-2+2sqrt7#