How do you solve #x^2 - 2x = 32#?

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Answer 1 · 2017-04-08T10:59:10

Solutuion : # x= 1 + sqrt 33 or x= 1 - sqrt 33 #

#x^2-2x=32 or x^2-2x +1 =32 +1 or (x -1)^2 =33 or x -1 = +- sqrt 33 #

or # x= 1 + sqrt 33 or x = 1 - sqrt 33 #

Solutuion : # x= 1 + sqrt 33 or x= 1 - sqrt 33 # [Ans]

Answer 2 · 2017-04-08T11:36:12

#color(blue)(x=1-sqrt33# or #color(blue)(x=1+sqrt33#

#x^2-2x=32#

#:.x^2-2x-32=0#

#:.x=(-b+-sqrt(b^2-4ac))/(2a)#

#:.x=-(-2)+-sqrt((-2)^2-4(1)(-32))/(2(1)#

#:.x=(2+-sqrt(4+128))/2#

#:.x=(2+-sqrt132)/2#

#:.x=2+-sqrt(2*2*3*11)/2#

#:.x=(2+-2sqrt33)/2#

#:.x=(2-2sqrt33)/2#

#:.x=(cancel2^color(blue)1(1-sqrt33))/cancel2^color(blue)1#

#:.x=1-sqrt33#

or:

#:.x=(2+2sqrt33)/2#

#:.x=cancel2^color(blue)1(1+sqrt33)/cancel2^color(blue)1#

#:.color(blue)(x=1+sqrt33#