$\Rightarrow m- 5\sqrt{m}=?$

I tried different approaches such as, substituting $\sqrt{m}=t$, squaring the given equations, substituting $\frac{m-m\sqrt{m}}{16}$ as $5$. Nothing seemed to work.

This is a multiple choice test with answer choices: $-26, -24, -20, 0, 24$

How can I solve this problem?

  • $\begingroup$ Are you sure you've copied the problem correctly? I don't see any sensible way to evaluate that expression...and doing it numerically doesn't appear to give a convenient answer. (of course, I might be missing something...) $\endgroup$ – lulu Apr 18 at 11:55
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    $\begingroup$ Something strange happens here. For $\sqrt{m}$ to exist, $m$ must be $\geq 0$. Then $80+ m\sqrt{m} =m \geq 80$. But then $m-m\sqrt{m}$ is negative. $\endgroup$ – liaombro Apr 18 at 12:02
  • $\begingroup$ @lulu. I checked again, the problem is copied correctly. $\endgroup$ – Eldar Rahimli Apr 18 at 12:02
  • $\begingroup$ @liaombro. I'm assuming there may not be a real solution for $m$, whereas $m-5\sqrt{m}$ would have a real result $\endgroup$ – Eldar Rahimli Apr 18 at 12:03
  • $\begingroup$ @liaombro Yes, you aren't looking at real numbers here. If we let $x=\sqrt m$, then we get the cubic $x^3-x^2+80$ which has roots $-4, \frac 12(5\pm \sqrt {55} \,i)$ $\endgroup$ – lulu Apr 18 at 12:05

I believe this is the intended solution.

From $m-m\sqrt{m} = 80$, we have $$(\sqrt{m}+4)(m-5\sqrt{m}+20) = 0.$$ Since clearly $\sqrt{m}\ne -4$, we must have $m-5\sqrt{m} + 20 = 0$. Thus, $$ m - 5\sqrt{m} = -20 $$

However, this ignores a point brought up by several other users: $m$ is not a positive real number. That means we're working with the complex $\sqrt{}$ function, which requires more care than the real one, as there are several identities true for the real $\sqrt{}$ function that aren't true of the complex one. It turns out that the algebra here is simple enough that those differences don't come up, but it's still an important distinction that shouldn't be glossed over.

  • $\begingroup$ Thanks, can you explain how did you factor the expression as$(\sqrt{m}+4)(m-5\sqrt{m}+20)$ $\endgroup$ – Eldar Rahimli Apr 19 at 5:21
  • $\begingroup$ @EldarRahimli I had Mathematica do it for me. I suppose one could also note that $m-m\sqrt{m} = (\sqrt{m})^2 - (\sqrt{m})^3 = 80 = 4^3 + 4^2$ and by inspection see that $(\sqrt{m}+4)$ must be a factor. $\endgroup$ – eyeballfrog Apr 19 at 15:15

There seems to be a problem with this question. The usual convention is that $\sqrt m$ refers only to the non-negative root, but in this case, the question is only solvable (even in the reals) if you allow for the negative root.

Let $\sqrt m = u$. The equation can be rewritten as $u^3 - u^2 + 80 = 0$, which can be shown (by sketching the curve for $y = u^3-u^2 = u^2(u-1)$ to guide the root search and the use of rational root theorem) to have a single real root at $u = -4$.

This leaves us in a bit of a quandary. If we adopted the convention we used above (take the negative root), the second expression works out to be $36$. But using the usual convention, we get $-4$. Perhaps both answers are meant to be stated.

By the way, the cubic also has a pair of conjugate complex roots, which weren't even considered in my answer. If you believe this is required, then you should work out the full solution of the cubic (not that hard to do once you've got the first factor of $u+4$ and divided by it to get a quadratic).

  • $\begingroup$ Thanks for the answer. This problem is supposed to be worked out without the use of calculators. Therefore, I believe I'm not expected to find the roots of the equation $u^3 + u^2 + 80=0$. Also, among the answer choices were only $-26, -24, -20, 0, 24$ $\endgroup$ – Eldar Rahimli Apr 18 at 12:18
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    $\begingroup$ @EldarRahimli It's not difficult to find the root $-4$ by hand using the method I outlined - I did. And none of those choices are correct. The complex roots give a complex result for the second expression (I just checked). Your question is likely just badly wrong. $\endgroup$ – Deepak Apr 18 at 12:22

I think that the problem is correct.

Let $x=\sqrt m$ to get $x^2-x^3=80$ the roots of which being $$x_1=-4 \qquad x_2=\frac{1}{2} \left(5-i \sqrt{55}\right)\qquad x_3=\frac{1}{2} \left(5+i \sqrt{55}\right)$$ $x_1$ must be discarded making $$m_2=-\frac{15}{2}-\frac{5 i \sqrt{55}}{2}\qquad m_3=-\frac{15}{2}+\frac{5 i \sqrt{55}}{2}$$

Now $$m_2-5 \sqrt {m_2}=m_2-5 x_2=-20$$ $$m_3-5 \sqrt {m_3}=m_3-5 x_3=-20$$


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