How do you solve #(4x – 3) ² – (2x – 1) ² = 0#?

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Answer 1 · 2016-05-15T17:32:18

#x=2/3, 1#

The given expression i of the form #a^2-b^2#

Where #a=4x-3# and #b=2x-1#

Now, #color(red)(a^2-b^2 = (a+b)(a-b))#

Then,
#(4x-3)^2-(2x-1)^2 =0#

#=> (4x-3+2x-1){4x-3-(2x-1)}=0#

#=> (6x-4)(4x-3-2x+1)=0#

#=> (6x-4)(2x-2)=0#

#color(red)( 6x-4=0)# or #color(blue)(2x-2=0)#

#color(red)( 6x=4)# or #color(blue)(2x=2)#

#color(red)( x=4/6)# or #color(blue)(x=2/2)#

#color(red)( x=2/3)# or #color(blue)(x=1)#