3x^2-24x+48=0 step by step please thanks

Question

jim_thompson5910 · Answer

3x^2-24x+48=0


3(x^2-8x+16)=0


3(x-4)^2=0


(x-4)^2=0


x-4=0


x=4

anonymous · Answer

ugh I dont get this ;/

jim_thompson5910 · Answer

which part is stumping you

anonymous · Answer

from 3(x-4)^2=0 to (x-4)^2=0, where did the 3 go? Thanks

anonymous · Answer

I know this is the right answer but I also don;t get the last part.. (x-4)^2 then the ^2 dissappears ??

jim_thompson5910 · Answer

With going from 3(x-4)^2=0 to (x-4)^2=0, I divided both sides by 3. 


On the left side, the 3 will cancel out and go away (since 3/3=1). On the right side, it will still remain zero since 0/3 = 0

anonymous · Answer

Ohh

jim_thompson5910 · Answer

With going from (x-4)^2=0 to x-4=0, I took the square root of both sides. This eliminates the square on the left side (since the square and square root "undo" each other). On the right side, the square root of 0 is 0. So that stays 0.

jim_thompson5910 · Answer

Yeah skipped a step or two, but hopefully that's clearer.

anonymous · Answer

Oh thanks so much I get it now, ur a good teacher (: