Simple square root question

But as he said, in maths exams, squaring a number you must put the positive and negative, regardless whether there is a +/- infront of the square root, unless this is something that comes up in university, we are taught the square root of any number can be +/-

This is definitely not something complex that only comes up at University.

The square root of 2 is 1.41...., not -1.41...
If I asked for +- root 2, I can have both answers, but if I only ask for the (positive) root, the answer is only 1.41...

If asked to solves X^2=2, then the solutions would be +-root 2. Note the use of the +- sign which indicates both roots are needed. If this +- sign is not present, only the positive root is taken.

Reply 41

Dalek1099

21

Original post by TenOfThem

It is not a "university" concept

It is the definition

It is scary how many teachers continue to teach this mis-conception

The scary thing is you can get into a lot of trouble in maths exams at AS by misusing sqrt(x) for example when differentiating things like y=xsqrt(x) if you get two values for dy/dx or y even at x=a then the examiner will drop marks and if sqrt(x)=x then a lot of implicit problems can be solved in incorrect easier ways.You must know sqrt(x)=+-x which is know as modulus x or abs(x) or you could lose easy marks.I have seen a lot of books get this wrong and it really angers me.The whole idea of why sqrt(x) can't give positive and negative answers is part of C3 on Functions with the fact that because y has more than one image then a restricted domain must be used.The same ideas for sqrt(x) must be used for things like arcsin(x) as well.

(edited 10 years ago)

Reply 42

TenOfThem

18

Original post by Dalek1099

The scary thing is you can get into a lot of trouble in maths exams at AS by misusing sqrt(x) for example when differentiating things like y=xsqrt(x) if you get two values for dy/dx or y even at x=a then the examiner will drop marks and if sqrt(x)=x then a lot of implicit problems can be solved in incorrect easier ways.You must know sqrt(x)=+-x which is know as modulus x or abs(x) or you could lose easy marks.I have seen a lot of books get this wrong and it really angers me.

No what you need to know is that there are 2 solutions to

x^2 = a

Reply 43

noobynoo

6

No but

\pm \sqrt{(\pm \hspace{0.1cm} x)^2} = \pm \hspace{0.1cm}|x| \hspace{0.1cm}

if x is a real number.

Reply 44

ihatebrownbread

I always thought that sqrt x meant all the possible values that if times by themselves make x.

Reply 45

BlueSam3

17

Original post by ihatebrownbread

I always thought that sqrt x meant all the possible values that if times by themselves make x.

This is incorrect.

Unparseable latex formula:

\sqrt

is a function,

\mathbb{R}_{\geq 0} \rightarrow \mathbb{R}_{\geq 0}

(and also on the rest of

\mathbb{C}

, but we're dealing with real valued square roots for now). That is: it returns a single (non-negative) value for each non-negative real number. It is often said that 1 has two square roots (and three cube roots, four fourth roots etc.), but only one of these is

\sqrt{1}

.

Simple square root question

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