I'm Not Even Sure Where 2 Begin?

[CalebsMomma]

Ok, so it says to find all real solutions to each equation...
here is the problem

x^2 + x+ (sqrt x^2 + x-2)=0

Please help?

 

[Aladdin]

Ok, so it says to find all real solutions to each equation...
here is the problem

x^2 + x+ (sqrt x^2 + x-2)=0

Please help?

Actually, it's a equation(one) -- not each . Type correct please.

Or your first problem is written un-correctly

And is it : \(\displaystyle ( \sqrt{x^2} + x - 2)\)

 

[CalebsMomma]

You are right there is just the one "equation" that I posted, but I just wrote what the directions for the whole set of problems, sorry for the confusion.

 

[Aladdin]

find all real solutions to each equation...

x^2 + x+ (sqrt x^2 + x-2)=0

\(\displaystyle x^2 + x + (\sqrt{x^2} + x - 2) = 0\)

\(\displaystyle x^2 + x + \sqrt{x^2} + x - 2 = 0\)

\(\displaystyle x^2 + 2x + \sqrt{x^2} - 2 = 0\)

\(\displaystyle x^2 + 2x + x - 2 = 0\)

\(\displaystyle x^2 + 3x - 2 = 0\)

 

[CalebsMomma]

I get it once you showed me, thanks, I forgot to square it times itself so I could add like terms! Thanks!!!

 

[Denis]

x^2 + x+ (sqrt x^2 + x-2)=0

Momma, you sure that's not x^2 + x + sqrt(x^2 + x - 2) = 0 ?

If so, then: x^2 + x = - sqrt(x^2 + x - 2)
square both side:
x^4 + 2x^3 + x^2 = x^2 + x - 2
x^4 + 2x^3 - x + 2 = 0

Have you been exposed to x^3 or x^4 yet?

 

[Deleted member 4993]

Ok, so it says to find all real solutions to each equation...
here is the problem

x^2 + x+ (sqrt x^2 + x-2)=0

Please help?

If it were:

x^2 + x+ sqrt (x^2 + x-2) = 0

transformed to (per Denis - which looks correct):

x[sup:29pj4a1e]4[/sup:29pj4a1e] + 2x[sup:29pj4a1e]3[/sup:29pj4a1e] - x + 2 = 0

Then this equation does not have a rational root. Graphing the function will show that it does not have any real root.

 

[mathematics]

hello,

Our teacher tought us , if the polynomial is in the degre of 3 and above you can group them then factorise.

(x)(x^3-1)(2)(x^3+1) = 0

is it correct ?

Sorry and what do you meam by rational roots please.

Thanks,

 

[Denis]

Our teacher tought us , if the polynomial is in the degre of 3 and above you can group them then factorise.
(x)(x^3-1)(2)(x^3+1) = 0
is it correct ?

No; quite wrong; better talk to your teacher.

x^4 + 2x^3 - x + 2 = 0 (no rational roots...as per Sir Khan)
If Momma made a sign typo and that should be:
x^4 + 2x^3 - x - 2 = 0 , then factorize:
x^3(x + 2) - (x + 2) = 0
(x^3 - 1)(x + 2) = 0
x^3 - 1 = 0: x^3 = 1; x = 1
or
x + 2 = 0: x = -2