Question

More about \[\large \sqrt{2x+1} = 2x - 5\]

The other night I solved this by factoring, not squaring the two sides. This got:

\[\large (\sqrt{2x+1} - 3)(\sqrt{2x+1} +2) = 0\]

One factor gives 4, which is the correct root. The other stops at

\[\large \sqrt{2x+1} = -2\]
which I think means an imaginary root. I'm calling the "ghost" of the extraneous root I would have introduced if like a sane person I had squared both sides. Does anyone have a better explanation or description of what is going on here?

Answer 1

factoring is the same as squaring , i mean when u factor u go back squaring again to get an answer ..

Answer 2

now , 6 is also solution but u couldnt get is by factor , right ?

Answer 3

@ikram002p I don't think 6 is a solution at any step of the process. It is the y (or f(x)) intercept. But that is not a solution for x.

Answer 4

sorry i means 3/2 xD

Answer 5

But 3/2 is not even on the curve of the problem as stated. It is another function assuming a minus sign on the radical. The red part of the graph is the given function.

Answer 6

well , as i said before factoring is the same as squaring , that means u dnt stop at
sqrt 2x+1 =-2
as long as u get an answer of x>=-1/2

Answer 7

say \(\large \sqrt{2x+1} = w\)
your factorization is equivalent to \(\large(w-3)(w+2)\)

Answer 8

Hey Lars, try squaring both sides

Answer 9

lars u have to consider after squaring
it would be like
|2x+1| =(2x-5)^2

Answer 10

which is a second degree polynomial, so the other root is not surprising
however it is extraneous to the original equation cuz you cannot factor it like that...

lets see what information the equation is missing when we factor it

Answer 11

ya that's what i was saying ikram

Answer 12

Whoops! I meant

\[ \large (2x+1) - \sqrt{2x+1} - 6 = 0 \]

Answer 13

sqrt(2x+1) = (2x-5)
=> 2x + 1 = (2x-5)^2

Answer 14

Yes. I know how to square both sides, but that introduces an extraneous root.

Answer 15

which is
2x + 1 = 4x^2 + 25 - 20x

Answer 16

You know the quadratic formula?

Answer 17

4x^2 - 22x + 24 = 0

Answer 18

which is 2x^2 -11x + 12

Answer 19

Now you can use the factor method here.

Answer 20

But I don't need the quadratic formula because it factors.

Answer 21

Yeah. You can use factor method. It will be quicker.

Answer 22

What could be quicker than this?

\[ \begin{align} \sqrt{2x+1} &= 2x - 5 \cr \sqrt{2x+1} - 2x + 5 &= 0 \cr 2x - \sqrt{2x+1} - 5 &= 0 \cr (2x+1) - \sqrt{2x+1} - 6 &= 0 \cr (\sqrt{2x+1} - 3)(\sqrt{2x+1} +2) &= 0 \end{align}\]

Answer 23

Here you need to work with roots. The best thing is to eliminate the roots first and solve the equation. You square both sides to eliminate the roots.

Answer 24

ok?

Answer 25

i see no hartm in this
sqrt 2x+1 =-2
:(

Answer 26

@LarsEighner , we're not done with solving :

\[\large \sqrt{2x+1} = -2 \]

solve this factor too, what do u get ?

Answer 27

@ikram002p The root of what number is a negative number?

Answer 28

It's right. If sqrt. ( 2x + 1) = -2
2x + 1 - (-2)^2

Answer 29

well , its ok if u got x >=-1/2

and as u said before using complex numbers , and it might lead u to a real root at the end

Answer 30

@ganeshie8 I think I get (3/2)i

Answer 31

no! square both the sides os the factor sqrt ( 2x + 1) = -2

Answer 32

which is
2x + 1 = 4

Answer 33

if thats correct, then there you have your other root :) (3/2i is not extraneous, it is a properr complex root)

what further explanation are you looking for ?

Answer 34

so x = 3/2

Answer 35

3/2i doesn't look correct though...

Answer 36

:'(

Answer 37

it won't be 3i/2.
Just 3/2

Answer 38

yes it would be 3/2 and extraneous, precicely because we have squared both sides haha!

Answer 39

yeah @ganeshie8

Answer 40

we always need to carry constraints when we do things like : \(x=a\implies x^2=ax \), constraint here is \(x \ne 0\)

Answer 41

um?

Answer 42

in the original equation, the inherent constraint is \(\large \sqrt{2x+1} \ge 0\)

Answer 43

yeah. i think

Answer 44

sqrt 2x+1 =-2

assume 2x+1 =u

then sqrt u= -2

sounds like complex to me , but also frusted xD

Answer 45

ok , i think the trick with domai range as ganesh stated

in both cases we would only get x=4 as an answer

Answer 46

by definition sqrt is nonnegative @ikram002p
are you trying to solve :
\[\large \sqrt{z} = -2\]

\(z\) some complex number
?

Answer 47

hehe i tried :3
but then it made no sense ,
since
|z|=4
not the same to
sqrt z =-2

Answer 48

shooo , i have nothing else to say

Answer 49

At
\[\large (2x+1) - \sqrt{2x+1} - 6 = 0\]

I am about to factor it as if it is:

\[\large a^2 - a - 6 = 0\]

where \(a = \sqrt{2x+1} \)

Did I go wrong here?

Answer 50

nope , it was correct lars

Answer 51

nothing wrong, only one constraint is missing... just add \(a\ge 0\) in your second line.

Answer 52

yess !

Answer 53

So when I get here

\[\large \large \sqrt{2x+1} = -2 \]

I cannot solve this without squaring it, but that creates the extraneous root?

Answer 54

well after trying i dont think there is extraneous root to solve it :o