Question

Would someone please check my answer.

Answer 1

Question :-
the product of all the sollutions of the equation
\[\huge (x-2)^{2} -3\left| x-2 \right| +2 =0 \]
Is?

Answer 2

My approach on solving this question:-

Substitute J as (x-2)
When (x-2) is positive
\[\huge J ^{2} -3J +2 =0\]
J=1,2
But J =(x+2)
so we get x= 3,0

When (x-2) is negative
\[\huge J ^{2} +3J +2 = 0\]
We again get J = -1 ,-2

x+2 =-1
x=-3

x+2 = -2
x=-4

SO product of all the sollutions

3*0*-3*-4 = 0

Answer 3

its zero
but ur solution :-\
i got 3,4 when positive
0,1 when negative

Answer 4

I have made a slight arithmetic error but ur answers seem incorrect too

Answer 5

I haven't made any error my answers are correct according to me i checked it thrice

Answer 6

0,1,3,4 are right - but u dont need to do all this
http://www.wolframalpha.com/input/?i=%28x-2%29%5E2-3%7Cx-2%7C%2B2%3D0

Answer 7

yeah mabye ^^
\((x-2)^2-3|x-2|+2=0\)
for x>=2
\((x-2)^2-3(x-2)+2=0\)
\((x^2-4x+4)-3x+6+2=0\)
\(x^2-7x+12=0\)
\((x-3)(x-4)=0\)
x=3,4

Answer 8

You're done at this step :
```
so we get x= 3,0
```

Answer 9

just conclude saying something like ; Since one solution is 0, the product of all solutions will be 0. QED.

Answer 10

and I noticed a small typo in ur work : J should be |x+2|

Answer 11

Well yeah , it striked me actually when i was done with writing it

Answer 12

a^2 = |a|^2

Answer 13

\[(x-2)^{2} -3\left| x-2 \right| +2 =0 = |x-2|^{2} -3\left| x-2 \right| +2 =0\]

Answer 14

(x+2)^2 would always be positive for any value of x , So why to put modulus
Sorry for opposing but did u get my point

Answer 15

Yes, cuz you want to turn it into a manageable quadratic

Answer 16

Yes

Answer 17

whats the question again ?

Answer 18

which question?

Answer 19

you're asking something right ?

Answer 20

No :)

Answer 21

lol sure ?

Answer 22

Yes xd

Answer 23

you're kindof saying x^2 is always positive so why to putting the mod is unnecessary, right ?

Answer 24

pardon for the bad grammar - lack of slp :/

Answer 25

Yes

Answer 26

then you're right ! thats the reason the initial quadratic came w/o any absolute value bars. It wants you to put them and work it exactly as u worked above^

Answer 27

but its wrong to solve for j= x-2

Answer 28

yeah, j = |x-2| fixes everything

Answer 29

Okay..............