I know that, $$\sqrt{25}=5≠-5$$
But, in quadratic equation we write $$(2ax+b)^2=b^2-4ac$$
$$(2ax+b)=±\sqrt{b^2-4ac}$$
Why?? We must write $±$, and What is the mathematical mystery here?
Asked
Active
Viewed 111 times
1
-
2From $(-3)^2=9$ you cannot conclude that $(-3)=3$. – Andrés E. Caicedo Jul 05 '17 at 20:51
-
1$5^2$ and $(-5)^2$ both equal $25$. Similarly, $x^2 = (-x)^2$. Thus, when applying $\sqrt{}$ to both sides of an equation you need to consider both the positive and negative roots. – Mosquite Jul 05 '17 at 20:53
-
1No mystery, the solutions of $y^2=25$ are $y=\pm 5.$ – gammatester Jul 05 '17 at 20:55
-
I think I'm not satisfied.. – Soru Jul 05 '17 at 21:02
2 Answers
1
In $\mathbb{R}$ there are two numbers $x$ such that $x^2=a$ (for $a>0$), and these two numbers have opposite sign, but, by definition, the symbol $\sqrt{a}$ indicates only the positive one of such two numbers, so, if we want indicate expliciltly all the two we must write $ x= \pm \sqrt{a}$.
Emilio Novati
- 64,377
-
-
No, $\sqrt{25}$ is, by definition, the positive number $x$ such that $x^2=25$, so it is $5$. It is calle also the principla square root: https://en.wikipedia.org/wiki/Square_root – Emilio Novati Jul 05 '17 at 21:10
-
Why?? ı am sorry, because I dont understand realy.. $-5$ is not a real root of $\sqrt{25}$ ?? – Soru Jul 05 '17 at 21:12
-
Yes, $-5$ is a real square root of $25$ but it is not the principal (that is positive) square root. so it is $-5=-\sqrt{25}$ – Emilio Novati Jul 05 '17 at 21:14
-
What you are telling us is that you do not understand what $\sqrt{x}$ means. You have already been told that $\sqrt{x}$ is defined as "the positive number whose square is x". -5 is not a root of $\sqrt{25}$ because it is not positive. Again, there are two solutions to $x^2= a$. One of them is $\sqrt{a}$, the other is $-\sqrt{a}$. – user247327 Jul 05 '17 at 21:16
-
-
-
The symbol $\sqrt{a}$ idicates only the positive root of $a$, so if we want the negative one we have to write $-\sqrt{a}$ and, if we want both the roots we write $\pm \sqrt{a}$. – Emilio Novati Jul 05 '17 at 21:25
-
-
-
0
You need to be careful as if we treat $\sqrt{x}$ as a function then we conventionally assume the value is $\ge0$ (there can't be two values as it wouldn't be a function at all), however $\sqrt{x}$ (or in general e.g. $\sqrt[n]{x}$ in $\mathbb{C}$) may denote the whole set of numbers such that each of them squared (raised to the $n$th power respectively) equals $x$, then it is obviously not a function.
Theta
- 887