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I'm reading for my matriculation exams and I now face a really odd problem. It all comes down to me overthinking about/doubting my equation solving , so I would like some clarification on second degree equations, since it's so long from when these ideas were introduced to me.

Say there's:

$$x^2 = a$$

What is the process via which all my books come to the conclusion that $x=±\sqrt{a}$ ?

or say for example $0<y<3$ and $$3 \cos^2(x) = y$$

I sort of "know" that you get $\cos(x) = \pm\sqrt{\frac{y}{3}}$.

But how does one arrive at that conclusion? All my math books' examples just "automatically" remove the exponent $2$ and add $\pm$, and so did I until now when I really started to think about it...

Searching the Internet I've only found an explanation that you square root both sides, and that removes the exponent leaving just $x$ or $\cos(x)$ respectively (shouldn't it be $|x|$ and $|\cos(x)|$?), and meanwhile the other side gets $\pm$ because apparently when you take the square root it can be the positive or the negative root (but didn't we take the positive root of both sides??).

Meanwhile my math book just straight up says that the solution of an equation of the form $x^2 = a$ is $x = \pm\sqrt{a}$ and leaves it at that. Am I correct to assume that this "property" is true whatever $x$ and $a > 0$ may be? (for example $x = y^2 + 3$ and $a = \ln(z)$ would yield $y^2 + 3 = \pm\sqrt{(\ln(z)}$)

So in case my question wasn't clear, how does one remove the exponent in these kind of equations. Do you square root both sides, or are you just supposed to "see" what the expression being squared equals to?

I know the question may seem trivial but it's annoying me to no end (I tend to not just "accept" the rules given, I usually want to see how they're derived)

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  • $\begingroup$ We call "square root of $a$" the number $b$ such that $b \times b=a$. $\endgroup$ – Mauro ALLEGRANZA Feb 26 '18 at 15:09
  • $\begingroup$ Due to the fact that also $(-b) \times (-b)=a$, we have two values for the equation: $x^2=a$, i.e. $x= + \sqrt a$ and $x= - \sqrt a$. $\endgroup$ – Mauro ALLEGRANZA Feb 26 '18 at 15:10
  • $\begingroup$ Are you asking how we know that the equation $x^2=a$ has a solution? $\endgroup$ – Lee Mosher Feb 26 '18 at 15:18
  • $\begingroup$ Well, I'm more or less asking how, in practice, do you solve equations with an exponent. Do you square root both sides, or something else? (for example just realize that since this expression is being squared, it therefore must be equal to the ±square root(a) $\endgroup$ – punctuationisimportant Feb 26 '18 at 15:23
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$\sqrt a$ is defined as the positive number such that $\sqrt a^2 = a$. (Clearly $-\sqrt a$ also has this property, but isn't positive.) So it's not a conclusion so much as "We know that this number exists, and since we're going to keep referring to it we need to give it a name. Let's call it $\sqrt a$."

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The equation $x^2=a$ is only solvable if $a\ge0$. t Also, $\sqrt a$ is defined if and only if $a\ge 0$.

For $a\ge0$, we have $(\sqrt a)^2=a$ and $(-\sqrt a)^2=a$, so $x^2=a$ means $x^2=(\sqrt a)^2=(-\sqrt a)^2$

This implies $x^2-(\sqrt a)^2=x^2-(-\sqrt a)^2=0$ $\Leftrightarrow (x-\sqrt a)(x+\sqrt a)=0$

We can jump to the conclusion that $x=\sqrt a$ or $x=-\sqrt a$.

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