1. ## solution of equation

Consider the following equation,

$x^{2}+\sqrt{x}-2=0$
$(\sqrt{x}+2)(\sqrt{x}-1)=0$
$\sqrt{x}=1$ or $\sqrt{x}=-2$
For the equation $\sqrt{x}=-2$, can I say that there is "no real solution".
Thus $x=1$

My teacher mark me wrong for the part "no real solution", she wrote "wrong reason for rejection".
Can someone please enlighten me? Am I correct?

2. Originally Posted by acc100jt
Consider the following equation,

$x^{2}+\sqrt{x}-2=0$
$(\sqrt{x}+2)(\sqrt{x}-1)=0$

Mr F says: One of the above two lines is wrong because they are not consistent with each other (expand the second line out to see this).

$\sqrt{x}=1$ or $\sqrt{x}=-2$
For the equation $\sqrt{x}=-2$, can I say that there is "no real solution".
Thus $x=1$

My teacher mark me wrong for the part "no real solution", she wrote "wrong reason for rejection".
Can someone please enlighten me? Am I correct?
Until you resolve the above problem (in red) nothing more can be done.

3. sorry, my mistake. The equation should be $x+\sqrt{x}-2=0$

4. Originally Posted by acc100jt
sorry, my mistake. The equation should be $x+\sqrt{x}-2=0$
$x+\sqrt{x}-2=(\sqrt{x}+2)(\sqrt{x}-1)$ is correct.
But I think that you need to ask your teacher the question.
It seems to be her preference which we have no way of knowing.

5. the fatorisation is correct. It is the part when I say "no real solution" that she marked me wrong

6. Originally Posted by acc100jt
the fatorisation is correct. It is the part when I say "no real solution" that she marked me wrong
I understand the OP.
But none of us here is likely to be your teacher.
That means we do not know why she did not like you answer.
7. When you said "no real solution" did you make it clear that you were only talking about the " $\sqrt{x}= -2$" term? Because x= 1 is clearly as solution and you get that from the other factor, $\sqrt{x}- 1= 0$.