Question

for a<0 determine all solution of the equation x^2-2aIx-aI-3a^2=0
I I symbol is modulus sign

Answer 1

@divu.mkr pls..hlp me in dis

Answer 2

\[x ^{2}-2a \left| x-a \right|-3a ^{2}=0\]

Answer 3

consider two cases x-a>0 x-a<0..btw you know how to solve mods?

Answer 4

keep in mind if x-a<0,x<a i.e.,x<0

Answer 5

consider:
\[x ^{2}-2a(x-a)-3a ^{2}=0\]and
\[x ^{2}-2a (a-x)-3a ^{2}=0\]

Answer 6

yaa i m solving it @amistre64

Answer 7

i m getting 2 values of x as a+-root 2a

Answer 8

\[x ^{2}-2a(x-a)-3a ^{2}=0\to x^2-2x-a^2=0\]
\[x = 1\pm\sqrt{1+a^2}\]
\[x^{2}-2a(a-x)-3a ^{2}=0\to x^2+2x-5a^2=0\]
\[x = -1\pm\sqrt{1+5a^2}\]

Answer 9

plug them back into the original setup to see if they all work :)

Answer 10

wich original setup are u talking abt

Answer 11

at the startof your post you were considering:\[x^2-2a|x-a|-3a^2=0\]

Answer 12

yaa right so do u want me to put the values in dis one

Answer 13

of course, it is the only one that we actually care about. Splitting it allows us to find critical values to test out.

Answer 14

donn't u think it wud be tedious process

Answer 15

lol, math isnt all speed and quickness

Answer 16

if theres another way to test them out .. then by all means give it a shot :)

Answer 17

im not getting a general result, but only specific cases for a value of a; like a=0, 1, 2sqrt2-3

Answer 18

answer is (1-root 2)a,(-1+root 6)a

Answer 19

hmm, accursed moduluses :)

Answer 20

-2ax ... i dropped an a :/

Answer 21

\[x ^{2}-2a(x-a)-3a ^{2}=0\]\[
x^2-2ax-a^2=0\]and +2ax for the other setup

then work it thru ....

Answer 22

@samigupta8 ur hot

Answer 23

@RadEn pls..hlp

Answer 24

oh, it's what amistre said
\[x^2-2a(x-a)-3a^2=0 \hspace{55px}x^2-2a(a-x)-3a^2=0\]
\[ x^2-2ax + 2a^2-3a^2=0 \hspace{55px}x^2-2a^2+2ax-3a^2=0\]
\[ x^2-2ax-a^2=0 \hspace{95px}x^2+2ax-5a^2=0\]
\[x=\frac{2a±\sqrt{4a^2+4a^2}}{2}\hspace{85px}x=\frac{-2a±\sqrt{4a^2+20a^2}}{2}\]
\[x=a±a\sqrt{2} \hspace{115px} x= -a ± a \sqrt{6} \]

Answer 25

i got that values alraedy

Answer 26

the after process is remaining

Answer 27

those are the answers though/...

Answer 28

I don't follow....

Answer 29

plggin them back in ?

Answer 30

yaa i m doing dat...

Answer 31

hmph.

Answer 32

\[x^2-2a \left| x-a \right|-3a^2=0\]
putting value as \[a+\sqrt{2}a\]
\[(a+\sqrt{2}a)^2-2a \left| a+\sqrt{2} a-a\right|-3a^2\]=0
\[a^2+2a^2+2\sqrt{2}a^2-2a(\sqrt{2}a)-3a^2\]=0
llhs=rhs=0

Answer 33

since a<0
|a| = -a

Answer 34

|a+root 2a-a| = |root 2 a| = -root 2a

Answer 35

-2a|a+root 2a-a| = + 2root 2 a^2
not -

Answer 36

okk...that means it wud be 4root 2 a^2 and not 0

Answer 37

yeah, now you will get 0 for (1-root 2) a
try it..

Answer 38

\[(a-\sqrt{2}a)^2-2a \left| a-\sqrt{2}a-a \right|-3a^2\]
\[a^2+2a^2-2\sqrt{2}a^2-2a(-\sqrt{2}a)-3a^2\]

Answer 39

it wud give me 0=0

Answer 40

and for -a-aroot6
\[(-a-a \sqrt{6})^2-2a \left| -a(2+\sqrt{6}) \right|-3a^2\]

Answer 41

how to solve dis now
shall i consider - sign or not

Answer 42

|-a*(...)| = -a (...)

Answer 43

|-a (2+sqrt 6)| = (2+sqrt 6) |-a| = (2+sqrt 6)(-a)\
as a<0

Answer 44

yeh dekho galat aa rha hai abb

Answer 45

it should not come =0 and its not coming, right ?

Answer 46

yaaaaaaaaaaa

Answer 47

yeh toh 8a^2+4a^2root6 aa rha hai

Answer 48

so, -a-aroot6 is not the solution
try other
it'll comes as 0

Answer 49

arrey maine wohi to try krke dekha tha

Answer 50

i hve considered x=-a-root 6 here

Answer 51

yes, and it did not =0 , right ?
so -a-aroot6 is not the solution

Answer 52

bt why it's given in answer

Answer 53

not actually....
,(-1+root 6)a means -a + root 6 a
which is given in answer as u said

Answer 54

yaa u r crrect...sry for my mistake i thought the minus sign was outside the bracket

Answer 55

no problem!
glad you finally solved it :)