What will be the value of $x$ in this equation? $$x^{3/2}+1=0$$
I am trying this question by shifting $1$ on the right hand side. So the above equation will become $$x^{3/2}=-1$$ So, the final value of $x$ will be $1$.
But the fact is that $x=1$ doesn't satisfy the above equation because $1+1$ is equal to $2$, not $0$.
Kindly help me out with this question?
Just to be clear, you did these manipulations?
$x^{\frac{3}{2}}+1=0$
$x^{\frac{3}{2}}=-1$
And then squared both sides? That is where your problem arises. When you squared both sides, you “lost information” specifically that it is a negative $1$ on the right side, not positive $1$.
Here is how I would do this:
Let $u=x^\frac{1}{2}$
Our equation is now:
$u^3=-1$
So we need a cube root of $-1$. There are three, and the simplest one fails. The other two are complex, and are most simply expressed as:
$e^{i\pi/3}$ and $e^{-i\pi/3}$
Now the final step is to square both of these to turn our $u$ values into $x$ values.
Therefore, $x= \{e^{2i\pi/3},e^{-2i\pi/3}\}$
