How do you solve #-12( x^{2} - 12) = - 9( 1+ 7x )#?

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Answer 1 · 2016-09-13T12:07:09

#x =1.8068#
#x =-7.0568#

At first glance, the fact that both sides are negative is uncomfortable. Multiply by #-1# for starters.

#-1xx-12( x^{2} - 12) = -1 xx- 9( 1+ 7x )#

#12( x^{2} - 12) = 9( 1+ 7x)" " larr # get rid of brackets

#12x^2 -144 = 9+63x" " larr# make equal to 0

#12x^2 -63x -153 = 0" "larr div 3#

#4x^2 +21x -51 =0#

This does not factorise, use the formula.

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

#x = (-21 +-sqrt(21^2-4(4)(-51)))/(2(4))#

#x= (-21 +-sqrt(441+816))/8#

#x = (-21 +sqrt1257)/8 #

#x =1.8068#

OR

#x = (-21 -sqrt1257)/8 #

#x =-7.0568#