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I ran into a Math problem in which I was told to solve like that:

\begin{align} x^ 2 - 25 &= 0 \\ x^2 & = 25 \\ x & = \sqrt{25}\\ x & = 5 \end{align}

I wonder what the logical and also intuictive explanation might be on removing the power of x and on the other hand, adding a square root to the other side.

Could you guys help me out on this? You see, even though I did solve the problem, I think it is only worthy if I understand the logical explanation behind it.

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(Assuming we already know that $x$ is positive.)

Say that we have come to the conclusion that $x^2=25$. This means that the number $x^2$ and the number $25$ are the same. If two numbers are the same, then their square roots are also the same.

The square root of $25$ is $5$, and the square root of $x^2$ is $x$. Therefore $x$ and $5$ are the same, and we write that as $x = 5$.

Every single rule you've learned about solving equations follow this line of thought. "The two numbers on either side of the $=$ are equal. Therefore they are still equal after we do the same thing to each of them" (in addition to "Swap one expression on one side of the $=$ for a simpler expression with the same value", those are the only two things you need to solve any equation ever).

In your case, for instance, the first step that you did (after assuming that $x^2-25 = 0$) was to add $25$ to both sides. The two sides were equal before you added $25$, therefore they must be equal after you added $25$. Then you swapped $x^2-25+25$ on the left-hand side with $x^2$, since that has the same value and is simpler, and the same with $0+25$ on the right side. (Most people do these in a single step, and some think of it as "moving the $-25$ over to the other side and swapping its sign", but you should, at the very least, keep in mind that these are the steps behind such a manouver.)

The trick to solving equations is to choose the things you do to both sides with care so that the expressions become simpler. For instance, to simplify $x^2$, the standard thing to do is to take square root (by definition of square root, that removes the $^2$, which simplifies things). If this also doesn't muck up the other side ($\sqrt{25}$ isn't too bad), then you actually go ahead and do that step.

As a non-example, consider $x^2 = x+2$. In this case, you could try to take square roots on both sides, and you would get $x = \sqrt{x+2}$. Yes, the left-hand side got nicer, but the right-hand side got worse. So that might not be the best step to take in that situation, even though it's entirely correct.

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  • $\begingroup$ Thank you, Arthur, what a brilliant mind! That was what I was looking for! $\endgroup$ – Matheus Minguini Jun 18 '18 at 13:03
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You have different ways to solve your problem.

$$x^2=25$$ $$ x^2-25=0$$ $$(x-5)(x+5)=0$$

$$ x= \pm5$$

Or,

$$x^2=25$$ $$ |x|=5 $$ $$ x= \pm5$$

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  • $\begingroup$ Mohammad Riaz-Kermani, I understood the logic behind the first way: You factored the first expression and it became a simple equation. I still cant understand the second way though, I mean, the logic behind $\endgroup$ – Matheus Minguini Jun 18 '18 at 12:56
  • $\begingroup$ When you take square root of $x^2$ you get $|x|$ not just $x$. That gives you two choices for $x$ which are $x=\pm 5$. If you know that $x$ is positive, then you choose $x=5$ $\endgroup$ – Mohammad Riazi-Kermani Jun 18 '18 at 13:57
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It should be

  • $x² - 25 = 0$
  • $x² = 25$
  • $\color{red}x = {\sqrt {25}}$
  • $x = 5$
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  • $\begingroup$ Thank you, Gimusi $\endgroup$ – Matheus Minguini Jun 18 '18 at 12:57
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The trick is to take the square root of the two members.

$$a=b$$ versus $$\sqrt a=\sqrt b.$$

This requires some care:

  1. if $a,b<0$, you may not take the square root;

  2. $\sqrt a=-\sqrt b$ is also true.

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  • $\begingroup$ Would something like "2. $\sqrt a = -\sqrt b$ could just as well be true" be more correct? $\endgroup$ – Arthur Jun 18 '18 at 13:12
  • $\begingroup$ @Arthur: indeed. $\endgroup$ – Yves Daoust Jun 18 '18 at 13:14
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What justifies that step is the exponentiation axiom $$(a^m)^n=a^{mn}.$$ If you agree with this, then from $x^2=25$, we obtain $(x^2)^{1/2}=25^{1/2}$, which gives $x=\sqrt{25}$, by another rule of exponentiation.

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I understand that where you are struct, if I am not wrong. Is the following your question?

$$ x^2 = 25 \implies x =5 ?$$

What is square root of $y$?

When you write $\sqrt{y} = \sqrt[2]{y}$, you mean that you are searching a number $x$ such that $x^2 = y$.

More generally, when we have $\sqrt[n]{y}$, it is nothing but a number $x$ such that $x^n = y$.

Having the above observations and notations,

$$x^2 = {25} \implies x = \sqrt{25} = \sqrt{5^2} =\sqrt{(-5)^2} $$ which evidently proves that $x = 5$ or $x = -5$.

If you are sure that $x$ is positive then $x = 5$.

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