Question

Help with the following ecuation.

|2x+5|-|x-4|=|x^2-1|

|a| means absolute value.
Thanks.

Answer 1

hi :) if you have tried it, tell me where you are stuck, we can take it from there....else it will take some time.

Answer 2

well, i've tried to elevate both sides to square in order to eliminate de absolute value, but then i get horrible 8 grade ecuation.
So i'm lost.

Answer 3

oh, when i square both sides, i just get a 4th degree eq.

Answer 4

\(|2x+5|-|x-4|=|x^2-1| \\ \sqrt{(2x+5)^2}-\sqrt{(x-4)^2}=\sqrt{(x^2-1)^2}\)
because \(|a|=\sqrt{a^2}\)
now square both sides...

Answer 5

Of course the firts time you'll get a 4th degree eq, but you need to square again to eliminate all the abs value.
\[(2x+5)^2-(x-4)^2+2*|2x+5||x-4|=(x^2-1)^2\]
If you expand that, you'll get
\[2*|2x+5||x-4|=x^4-7x^2-12x-40\]
So i think this method so much tricky.

Answer 6

if i square this :
\(|2x+5|-|x-4|=|x^2-1| \\ \sqrt{(2x+5)^2}-\sqrt{(x-4)^2}=\sqrt{(x^2-1)^2}\)

i get
\((2x+5)^2+(x-4)^2-2 \sqrt{(2x+5)^2}\times \sqrt{(x-4)^2}=(x^2-1)^2 \\(2x+5)^2+(x-4)^2-2 {(2x+5)}\times {(x-4)}=(x^2-1)^2 \)

Answer 7

hmmm.
why
\[2\sqrt{(2x+5)^2}*\sqrt{(x-4)^2}\]
Becomes
\[2(2x+5)(x-4)\]
It shouldn't be
\[2*\left| 2x+5 \right|\left| x-4 \right|\]
?

Answer 8

yes, sorry...i am incorrect.

Answer 9

i'll try to think of other ways...

Answer 10

instead of squaring, lets take 2 cases everytime,
\(x-4 =|2x+5|-|x^2-1| \quad or \quad x-4=|x^2-1|-|2x+5|\)
do this repeatedly

Answer 11

for first one
\(|2x+5|=x-4+|x^2-1| \implies 2x+5=x-4+|x^2-1| \\ or, \: 2x+5=4-x-|x^2-1|\)

Answer 12

getting what i am trying to do ?

Answer 13

Yes, and we'll get 8 possible combinations.

Answer 14

better than solving 8th degree equation...

Answer 15

it would take time to solve 8 quadratic equations and check whether that particular value is actually the root of original equation or not.....but i can't find an easier way....can you continue this way? or still need help with this method ?

Answer 16

This is better than squaring thinks.
But happens that some values don't match with the original ecuation.
Anyway, the answer it´s supposed to be x=0, and one combination provides this value.
So we have to test every results and discard.
I think i get it.
Thanks.

Answer 17

hmm...welcome ^_^