How do you solve #(8x-9)^2=6#?

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Answer 1 · 2017-02-27T18:23:59

#x = 1.431" and "x =0.819#

If you had an equation such as #x^2 = 25,#

you could solve it easily by taking the square root of both sides:

#x = +-sqrt25 = +-5#

The given equation is in exactly the same form:

#(8x-9)^2 =6" "larr# take the square root of both sides

#8x-9 = +-sqrt6" "larr# remember there are two roots

#8x = +-sqrt6 +9#

This will give us two answers:

#x = (+sqrt6+9)/8" and "x = (-sqrt6+9)/8#

#x = 1.431" and "x =0.819#