Solving $x^{3/2}+1=0$

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?

• Welcome to MSE. You should choose your tags carefully. What has this to do with linear-algebra? Commented Aug 6, 2023 at 4:45
• You probably will need complex numbers here. Commented Aug 6, 2023 at 4:47
• What complex value is possible? Commented Aug 6, 2023 at 4:58
• The topic is mainly of linear equations Commented Aug 6, 2023 at 4:59
• For future reference, even if this were a linear equation (which it is not), linear equations don't generally fall under the topic of linear algebra. Read the descriptions of tags before using them. Commented Aug 6, 2023 at 5:38

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, $$x= \{e^{2i\pi/3},e^{-2i\pi/3}\}$$

• If the question is (x^5/2 )+1=0 then how I will solve? Commented Aug 6, 2023 at 12:43
• You would need to find fifth roots of $-1$, and then square them. Commented Aug 6, 2023 at 15:03