Question

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

Answer 1

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

Answer 2

ugh I dont get this ;/

Answer 3

which part is stumping you

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

Ohh

Answer 8

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.

Answer 9

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

Answer 10

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