1

In as part of solving a question, the equation

$a-3\sqrt a-4=0$

is written into

$a^2-3a-4=0$

How is this done? Do you square everything in the equation? But in this case why are only the $a$ squared?

Thanks!

Spica
  • 291
  • 3
    I don't think that manipulation is correct; if we make a substitution $u^2=a$ then $u^2-3u-4=0$ but $u$ isn't $a$... (We can also write a quadratic in $a$, but it'd be different) – Milo Brandt Nov 08 '15 at 19:36
  • Simple substitution: set $b=\sqrt a$, whence $a=b^2$. The equation becomes $b^2-3b+4=0$. Then you can decide $b$'s name is $a$. – Bernard Nov 08 '15 at 19:38

1 Answers1

3

This is a trick, and you are probably confused that they changed the variable, but did not change the variable name.

I change the variable name, so the trick is to define $b=\sqrt a$.

Then $b^2 - 3b - 4 = 0$.

This is exactly your second equation, but where the $a$'s have been replaced by $b$'s. From your first equation $a-3\sqrt a - 4 =0$, when $b=\sqrt a$, you get $a = b^2$, so this is why the equation becomes $b^2-3b-4=0$.

Mankind
  • 13,170
  • Thank you! Later on in the question, it says that $-1=\sqrt (x+2)$ isn't possible, how so? Would it not be then $x=-1$? – Spica Nov 08 '15 at 19:52
  • @Spica To see this, you only need remember one thing about the square root function, and it is that the square root function only spits out non-negative values. So even if you plug $x=-1$ into the function, you get $\sqrt{x+2} = \sqrt{-1+2} = \sqrt{1} = 1\neq -1$. It is not possible for a square root to be $-1$, because $-1$ is negative. – Mankind Nov 08 '15 at 19:59
  • But would $\sqrt 1$ not be ±1 just like $\sqrt 4$ is ±2? – Spica Nov 08 '15 at 20:23
  • @Spica My guess is that in the situations you usually encounter, you don't think of $\sqrt{4}$ as $\pm 2$, but rather you have an equation like $x^2 = 4$, and you then rewrite it as $x = \pm\sqrt{4}$. Since $\sqrt{4}=2$, you would have that $x = \pm 2$. The safest thing would be to think of the square root as a genuine function, and functions always return only one value (so you can't have that $\sqrt{4}$ is both $2$ and $-2$). Let me know, if it doesn't make sense. – Mankind Nov 08 '15 at 21:27
  • I think I understand. Is there any way to distinguish where I should see $\sqrt 4$ as ±2 and where I should see $\sqrt 4$ as just 2? – Spica Nov 08 '15 at 21:36
  • @Spica I would say that you should always just regard $\sqrt{4}$ as $2$. If you have an equation like $x^2=4$, then you get two solutions, $x=\sqrt{4}$ and $x=-\sqrt{4}$. In both instances, you regard $\sqrt{4}$ as $2$. You are correct that in some branches of mathematics, you do think of $\sqrt{4}$ as having both of the values $2$ and $-2$, but then $\sqrt{x}$ is no longer a function. My guess is that in all of the applications of the square root that you come by, just think of the square root as returning only the positive root. – Mankind Nov 08 '15 at 22:12
  • Can you come up with any situation, where it makes sense to you to think of $\sqrt{4}$ as $\pm 2$? – Mankind Nov 08 '15 at 22:12
  • Yes, for example the quadratic formula $\frac{-b±\sqrt (b^2 -4ac)}{2}$, where having two answers for the square root would give the two roots of the equation. – Spica Nov 08 '15 at 22:27
  • @Spica yeah, if you had two answers for the square root, then you could write that formula as $\frac{-b+\sqrt{b^2-4ac}}{2a}$, with the understanding that $\sqrt{b^2-4ac}$ means both $+\sqrt{b^2-4ac}$ and $-\sqrt{b^2-4ac}$, just like you could think of $\sqrt{4}$ as meaning both $+2$ and $-2$. But this is very confusing. Instead, if you think of the square root as returning only a positive number, then you need to include both signs $\pm$ in the formula, as you did, e.g. $\frac{-4\pm\sqrt{4}}{2}$ means $\frac{-4\pm 2}{2}$, because $\sqrt{4}$ only equals $2$. – Mankind Nov 08 '15 at 22:57
  • Oh, oh I see. That makes sense. Thank you! – Spica Nov 08 '15 at 23:15