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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: qshyrn on October 23, 2009, 10:40:37 pm

Title: Back to the basics: ... what is Root 4?
Post by: qshyrn on October 23, 2009, 10:40:37 pm: i had some calculus problem in class (although the problem wasnt really calculus if u get what i mean lol)... so it started off: (this is the first time ive done latex forgive me if i did it wrong)

$\int_{1}^{4}(6x-3\sqrt{x})= \left [ \frac{6x^2}{2}-\frac{6x^(3/2)}{3} \right ]\begin{matrix} 4 \\ 1 \end{matrix}(please. someone. teach. me. how . to . do .the . proper .notation .on . latex!) =\left [ 3x^2-2x\sqrt x \right]\begin{matrix} 4 \\ 1 \end{matrix} =(3*16-8\sqrt4)-1=31 (taking . the. \sqrt 4 .as. +ve$
now... my teacher solves this and gets 31.. but then i ask her why you cant take a negative value for the root 4 ... and she couldnt answer!
im sorta confused... its sorta obvious for the problem that u should take root 4 as positive two ,but WHY CANT YOU TAKE IT AS -2
what is the value of root 4?
Title: Re: Back to the basics: ... what is Root 4?
Post by: /0 on October 23, 2009, 10:50:26 pm: Because by definition, $\sqrt{x}$ denotes the positive square root. The function $f(x) = \sqrt{x}$ is only positive.
If you are asked to evaluate $\sqrt{4}$ the answer is 2, but if you are asked to find the (edit: square) roots of 4, the answer is $\pm 2$ . The difference is in the definition.

Also, the latex for $\left[x^2\right]_0^1$ is "\left[x^2\right]_0^1"
Title: Re: Back to the basics: ... what is Root 4?
Post by: shinny on October 23, 2009, 10:51:52 pm: Quote from: qshyrn on October 23, 2009, 10:40:37 pm
and she couldnt answer!

Oh god.

Quote from: qshyrn on October 23, 2009, 10:40:37 pm
WHY CANT YOU TAKE IT AS -2
what is the value of root 4?

Try graphing $y=\sqrt{x}$ , and this might explain some of it. But basically, the formal definition of $\sqrt{x}$ is that you only take the positive root. The negative root must be denoted by $-\sqrt{x}$ , and both would hence be $\pm \sqrt{x}$ .

EDIT: Beaten by /0.
Title: Re: Back to the basics: ... what is Root 4?
Post by: qshyrn on October 23, 2009, 10:52:17 pm: Quote from: /0 on October 23, 2009, 10:50:26 pm
Because by definition, $\sqrt{x}$ denotes the positive square root. The function $f(x) = \sqrt{x}$ is only positive.
If you are asked to evaluate $\sqrt{4}$ the answer is 2, but if you are asked to find the roots of 4, the answer is $\pm 2$ . The difference is in the definition.

Also, the latex for $\left[x^2\right]_0^1$ is "\left[x^2\right]_0^1"
hmm ok thanks.. ive always thought the root symbol means 'something times something' equals the number inside the root
Title: Re: Back to the basics: ... what is Root 4?
Post by: shinny on October 23, 2009, 10:53:31 pm: Quote from: qshyrn on October 23, 2009, 10:52:17 pm
Quote from: /0 on October 23, 2009, 10:50:26 pm
Because by definition, $\sqrt{x}$ denotes the positive square root. The function $f(x) = \sqrt{x}$ is only positive.
If you are asked to evaluate $\sqrt{4}$ the answer is 2, but if you are asked to find the roots of 4, the answer is $\pm 2$ . The difference is in the definition.

Also, the latex for $\left[x^2\right]_0^1$ is "\left[x^2\right]_0^1"
hmm ok thanks.. ive always thought the root symbol means 'something times something' equals the number inside the root

Yeh, it's important to just recognise that it is a function, and hence cannot return two answers.
Title: Re: Back to the basics: ... what is Root 4?
Post by: qshyrn on October 23, 2009, 11:00:57 pm: Quote from: shinny on October 23, 2009, 10:51:52 pm
Quote from: qshyrn on October 23, 2009, 10:40:37 pm
and she couldnt answer!

Oh god.

Quote from: qshyrn on October 23, 2009, 10:40:37 pm
WHY CANT YOU TAKE IT AS -2
what is the value of root 4?

Try graphing $y=\sqrt{x}$ , and this might explain some of it. But basically, the formal definition of $\sqrt{x}$ is that you only take the positive root. The negative root must be denoted by $-\sqrt{x}$ , and both would hence be $\pm \sqrt{x}$ .

EDIT: Beaten by /0.
lol 'oh god' .. . yes this was a very surprising mistake by my teacher, she never gets things wrong and is VERY good at explaning things (AND SHE IS AN ASIAN TOO)

anyway i sorta cleared it up afterschool with my dad but i didnt trust him 100% so i decided to ask on vn.
Title: Re: Back to the basics: ... what is Root 4?
Post by: TrueTears on October 23, 2009, 11:03:28 pm: Yes, I think /0 makes a really good point, there is a subtle difference between finding the square roots of something and to find $\sqrt{x}$ .

This happens quite a lot in complex numbers.
Title: Re: Back to the basics: ... what is Root 4?
Post by: kamil9876 on October 23, 2009, 11:12:41 pm: May be a bit of topic, but sort of related to some cool shit I was thinking about today: Prove that if a polynomial with only rational coefficients, $p(x)$ , has $p(a+\sqrt{2})=0$ where $a$ is rational, then $p(a-\sqrt{2})=0$ .
Title: Re: Back to the basics: ... what is Root 4?
Post by: NE2000 on October 24, 2009, 03:55:29 pm: Quote from: qshyrn on October 23, 2009, 10:40:37 pm
i had some calculus problem in class (although the problem wasnt really calculus if u get what i mean lol)... so it started off: (this is the first time ive done latex forgive me if i did it wrong)

$\int_{1}^{4}(6x-3\sqrt{x})= \left [ \frac{6x^2}{2}-\frac{6x^(3/2)}{3} \right ]\begin{matrix} 4 \\ 1 \end{matrix} =\left [ 3x^2-2x\sqrt x \right]\begin{matrix} 4 \\ 1 \end{matrix} =(3*16-8\sqrt4)-1=31 (taking . the. \sqrt 4 .as. +ve$
now... my teacher solves this and gets 31.. but then i ask her why you cant take a negative value for the root 4 ... and she couldnt answer!
im sorta confused... its sorta obvious for the problem that u should take root 4 as positive two ,but WHY CANT YOU TAKE IT AS -2
what is the value of root 4?

Essentially:

Solve: $x^{2} = 4$
$x = \pm 2$

Solve: $x = \sqrt {4}$
$x = 2$

This leads to another concept though and that is creating solutions. Essentially when you square something you can create an extra solution. Obviously this is not what you want. So for example if I had $x = \sqrt {4}$ and decided to square both sides I get $x^{2} = 4$ and then I finally get $x = \pm 2$ which is incorrect. So beware!