How do you solve #-7sqrt(3x-8)+3=-4x+7#?
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You will have to use the quadratic method after simplifying and moving all terms to one side of the equation.
Original Equation:
#-7\sqrt(3x-8)+3=-4x+7#
Subtract 3 from both sides...
#-7\sqrt(3x-8)\cancel(+3)\cancel(-3)=-4x+7-3#
#-7\sqrt(3x-8)=-4x+4#
Divide both sides by -7...
#\frac{\cancel(-7)\sqrt(3x-8)}{\cancel(-7)}=(-4x+4)/(-7)#
#\sqrt(3x-8)=(-4(x-1))/(-7)#
Square both sides to get rid of square root...
#(\sqrt(3x-8))^2=((-4(x-1))/(-7))^2#
#3x-8=(16(x-1)^2)/(49)#
Multiply both sides by 49...
#49(3x-8)=\cancel(49)\cdot(16(x-1)^2)/(\cancel(49))#
#147x-392=16(x-1)^2#
Expand the square #(x-1)^2#, then simplify...
#147x-392=16(x^2-2x+1)#
#147x-392=16x^2-32x+16#
Move all terms to one side and solve the polynomial.
#0=16x^2-32x-147x+16+392#
#0=16x^2-179x+408#
Use the quadratic formula to solve for x.
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(179+-sqrt(-179^2-4*16*408))/(2*16)#
Simplify.
#x=(179+-sqrt(32041-26112))/(32)#
#x=(179+-sqrt(5929))/(32)#
#x=(179+-77)/(32)#
#x=(256)/(32)# AND #x=(102)/(32)#
#x=8# AND #x=51/16#
