How do you solve the quadratic #4x^2-12x=16# using any method?

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Answer 1 · 2016-09-18T09:53:39

#x=4#
#x=-1#

Given -

#4x^2-12x=16#

#4x^2-12x-16=0#

#4x^2+4x-16x-16=0#

#4x(x+1)-16(x+1)=0#

#(4x-16)(x+1)=0#

#4x-16=0#

#x=16/4=4#

#x+1=0#

#x=-1#

Answer 2 · 2016-09-18T10:03:42

#x=4# or #x = -1#

#4x^2-12x=16#

#4x^2 -12x-16 =0" "larr"div4#

#x^2-3x-4=0" "larr# factorise

Find factors of 4 which subtract to give 3.

#(x-4)(x+1) =0#

If # x-4 = 0, " then " x = 4#

If #x+1 = 0, " then " x = -1#