Why is $u^{1/2}=-5$ not a solution?

Question

\begin{align} u+2u^{1/2}-15&=0\\ (u^{1/2}+5)(u^{1/2}-3)&=0\\ \end{align}

why is $u^{1/2}=-5$ not regarded as a solution, even though $(u^{1/2})^2=(-5)^2=(25)$? Is it because $\sqrt{u}$ can never be negative?

Answer 1

Try it:

If $u^{\frac 12} = -5$ (which is not a conventionally acceptable definition)

The $u = 25$.

The $u + 2u^{\frac 12} -15 = $

$25 + 2*(25)^{\frac 12} - 15 = $

$25 + 2*5 - 15=$ (I'm sorry... if $25^{\frac 12}$ is supposed to have a single value it must be by convention the positive value)

$25 + 10-15 =20 \ne 0$.

Now had the question been: Let $k$ be a number so that $k^2 = u$ then solve for $u$ where

$u + 2k - 15 = 0$

So $k^2 + 2k -15 = 0$

$(k+5)(k-3) = 0$

So $k=-5,3$ and $u =k^2 = 25,9$ would be fine.

Another way of putting that question would be:

Solve for $u$

$u + 2(\pm u^{\frac 12}) -15 = 0$

$(\pm u^{\frac 12} +5)(\pm u^{\frac 12} -3) =0$

So we have $\pm u^{\frac 12} = -5$ so $\pm = -$ and $-u^{\frac 12} = -5$ and $u =25$

Or $\pm u^{\frac 12} = 3$ so $\pm = +$ and $u^{\frac 12} = 3$ and $u=9$.

That's a weird way of looking at it but it works.

$u \pm 2 u^{\frac 12} -15$ has solutions whent $u =25$ and the $\pm $ is minus. And when $u = 3$ and the $\pm$ is positive.