3(x-4)^2-5(x-4)=12 find iti roots by factorisation
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3(x-4)²-5(x-4)=12
3(x²-8x+16)-5(x-4)=12
3x²-24x+48-5x+20=12
3x²-24x-5x+48+20-12=0
3x²-29x+56=0
3x²-8x-21x+56=0
(3x²-8x)-(21x+56)=0
x(3x-8)-7(3x-8)=0
(3x-8)(x-7)=0
(3x-8=0)(x-7=0)
x=8/3 and x=7
Therefore, it's roots are 8/3 and 7.
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