Question

Help needed please :
|x-1| < 2|x+2|

Answer 1

Have you tried converting the absolutes to squares?

Answer 2

yes i did @lightchaste, but do i have to squares |x+2| or the whole 2|x+2|?

Answer 3

not 2. coz it's not in mod.

Answer 4

Just the |x+2| :)

Answer 5

okay, so i've got the answers in decimals. but apparently the answers i the answer scheme is in integers

Answer 6

What's the next step? @lightchaste

Answer 7

What's the answer?

Answer 8

the answer is x<-5 and x>-1

Answer 9

Correct.

Answer 10

ys thats what twolf also says
http://www.wolframalpha.com/input/?i=%7Cx-1%7C+%3C+2%7Cx%2B2%7C

Answer 11

okay so @saifoo.khan, how did it get to be -5 and -1 bcs i seem to be getting decimals while doing the quadratic equation.

Answer 12

i myself dont know how to solve after solving the square roots.. hehe

Answer 13

thats why, right. haha. but the answer is correct?

Answer 14

Yes.

Answer 15

did u get \[x ^{2}+10x +7\]?

Answer 16

i have no idea how u got that. ;D

Answer 17

I did. But try looking at the wolfram link

Answer 18

\[\sqrt{(x-1)^{2}}<2\sqrt{(x+2)^{2}}\]

Answer 19

haha this is what i did, do correct me if i'm wrong\[\left( x-1 \right)^{2} < 2 \left( x+2 \right)^{2}\]\[x ^{2}-2x +1 < 2x ^{2}+8x+8\]\[x ^{2}+10x+7 >0\]then i'm all lost

Answer 20

Where'd the square roots go now?

Answer 21

ya u should not take away the square root!!

Answer 22

isn't square root and another square root can be cancel off?

Answer 23

Yep, and it kinda gives a clue of where the -5 came from but yeah other than that I'm not really sure.

Answer 24

ok wait, so the -5 came from where again?

Answer 25

so how do we know how to put that square root and when not to?

Answer 26

since you can cancel off the square roots and the squares its like a normal equation of \[x-1<2(x+2)\]

Not sure Kiki :( kinda confused with some of the rules taught by Mr. Nathan

Answer 27

i dont really understand what the wolfram site says hahaha @lightchaste

Answer 28

okay got it! thanks everyone :)

Answer 29

Wait. how?

Answer 30

i think there's something to do with the square root thing. \[x-1<2(x+2)\]\[x-1<2x+4\] and \[x-1<-4-2x\]But still the answer is not that accurate. right?

Answer 31

\[x-1<2x + 4\]
\[-x<5\]
\[x>-5\]

But this isn't accurate either :(

Answer 32

@fatinatikah You're wrong at first,
\[(x−1)^2<(2(x+2))^2\]
\[x^2-2x+1<4(x^2+4x+4)\]
\[x^2-2x+1<4x^2+16x+16\]
\[3x^2+18x+15>0\]
3(x^2+6x+5)>0
3(x+1)(x+5)>0
x=-1 , x=-5

Ans
x<-5 , x>-1

Answer 33

:D Awesome! so you have to include the number 2?

Answer 34

yes, you square both sides

Answer 35

so it isn't just basically converting the absolutes? And how do we determine whether the absolute needs to be converted into squares or with a square root?

Answer 36

just square them

Answer 37

owh okay :)