Question

3l2x+4l-1=11

Answer 1

3l2x+4l-1=11
3l2x+4l =11+1
3l2x+4l =12
l2x+4l = 12/3
l2x+4l = 4

Answer 2

can u continue it ?
dont forget |f(x)| = sqrt(f(x)^2)

Answer 3

after @tanjung
2|x+2|= 4
or |x+2|=2
or x+2=2 or x+2=-2
i.e x=0,-4

Answer 4

im really confused!

Answer 5

\[|2x+4|=4\]\[(2x(2x+4)+4(2x+4))=4\]\[4x ^{2}+8x+8x+16=4\]\[4x ^{2}+16x+16-4=4-4\]\[4x ^{2}+16x+12=0\]\[x ^{2}+16x+48=0\]\[(x=12)(x+4)=0\]\[x _{1}=-12\]\[x _{2}=-4\]

Answer 6

but -12 is not the solution here

Answer 7

none of my options involve -12

Answer 8

what if -4?

Answer 9

-4 is the solution here

Answer 10

you can choose one of the answer

Answer 11

is -0 an answer too?

Answer 12

u can check 0 is also one of the solution

Answer 13

no, only -4 or -12

Answer 14

if they have mentioned one answer then u can choose from 0 or -4