Radical simplification question

[i-love-physics]

Homework Statement

Write the expression in simplest radical form, if a radical appears in the denominator, rationalize the denominator

Homework Equations

\sqrt{9x^2-6x+1}

The Attempt at a Solution

the answer in the book says the answer is 3x - 1

I do not understand this, because i thought the rule was that the sum of radicals cannot be simplified

can someone please explain to me how this happened, it would be greatly appreciated.

 

[robphy]

Can you simplify the expression inside the square root symbol?
Strictly speaking, I don't think the book's answer is correct.
(By the way, this is not the sum of radicals... it's the radical of a sum.)

 

[i-love-physics]

i can simplify the expression inside the square root
this will give me

\sqrt{3x(3x-2)+1}


doesnt seem to get me any closer to the books answer


if the books answer is wrong, what is the real answer?

 

[symbolipoint]

Big HINT: the expression under the radical is a squared binomial.

Fill in the missing parts:
(3x )(3x )

 

[i-love-physics]

Big HINT: the expression under the radical is a squared binomial.

Fill in the missing parts:
(3x )(3x )

beautiful hint symbolipoint, i just figured it out thanks to you!

Thanks heaps mates, that was a good hint!, your a genious

there's some tricky bloody questions in maths sometimes I tell you mate.

 

[Gib Z]

Lol yup bloody tricky :) You could have checked if the answer in the book was right or wrong by squaring it. If the square of the books answer was 9x^2-6x+1, then its correct isn't it?

Well as robphy said already, not completely correct. It should have 2 solutions, the positive as well as negative one.

 

[Curious3141]

Well as robphy said already, not completely correct. It should have 2 solutions, the positive as well as negative one.

Not really. Robphy is correct, but not in the sense you mean. There is only one root implied by the radical sign, and that's the positive root.

So the answer should be |3x-1|, valid for all real x.

I had tried to post that before but got caught in the downtime.

 

[Rhythmer]

HINT:

\sqrt{9x^2-6x+1}

= \sqrt{9x^2-(3+3)x+1}

= \sqrt{9x^2-3x-3x+1}


Try to complete it yourself.

 

[pooface]

the back of the book is correct. 3x-1 is the right answer.

check this out. let us deal with the expression in the radical first.

9x^2 - 6x +1. As hinted by the fellow above it is a squared binomial.

so we FACTOR. 9x^2-6x+1=(3x -1)(3x-1)

So =(3x-1)^2.

hmmmmm so the square root of that equaaaaaaaals? cmon you can do this now. two ways to solve this. a squared term inside a SQUARE root? HMM!

or you do know that square root is actually exponent 1/2. try on your calculator 64^.5 you will get 8!

so if square root is 1/2, use your exponent laws to simplify your expression.

 

[i-love-physics]

the back of the book is correct. 3x-1 is the right answer.

check this out. let us deal with the expression in the radical first.

9x^2 - 6x +1. As hinted by the fellow above it is a squared binomial.

so we FACTOR. 9x^2-6x+1=(3x -1)(3x-1)

So =(3x-1)^2.

hmmmmm so the square root of that equaaaaaaaals? cmon you can do this now. two ways to solve this. a squared term inside a SQUARE root? HMM!

or you do know that square root is actually exponent 1/2. try on your calculator 64^.5 you will get 8!

so if square root is 1/2, use your exponent laws to simplify your expression.

Yeh i know, i figured it out as soon as symbolipoint gave me the hint

 

[robphy]

the back of the book is correct. 3x-1 is the right answer.

check this out. let us deal with the expression in the radical first.

9x^2 - 6x +1. As hinted by the fellow above it is a squared binomial.

so we FACTOR. 9x^2-6x+1=(3x -1)(3x-1)

So =(3x-1)^2.

hmmmmm so the square root of that equaaaaaaaals? cmon you can do this now. two ways to solve this. a squared term inside a SQUARE root? HMM!

or you do know that square root is actually exponent 1/2. try on your calculator 64^.5 you will get 8!

so if square root is 1/2, use your exponent laws to simplify your expression.

So, sqrt(x^2)=x ??
I can see that sqrt(64)=8...

but suppose my x was equal to -8...
then sqrt( (-8)^2) = sqrt (64) = 8 , which is not x, but -x, ... in my case... as tried on my calculator.

The calculator calculates |x|. The square root of x^2, sqrt(x^2), is |x|.

http://mathworld.wolfram.com/SquareRoot.html says
"Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of 9 are -3 and +3, since (-3)^2==(+3)^2==9. Any nonnegative real number x has a unique nonnegative square root r; this is called the principal square root and is written r==x^(1/2) or r==sqrt(x). For example, the principal square root of 9 is sqrt(9)==+3, while the other square root of 9 is -sqrt(9)==-3. In common usage, unless otherwise specified, "the" square root is generally taken to mean the principal square root."

So, as I hinted at above [and stated explicitly by Curious3141], the real answer to the original problem (following "common usage" and "my calculator") is |3x-1| ... otherwise, use the pair: (3x-1), -(3x-1) [as stated explicitly by Gib Z].

[Note that 9x^2-6x+1 can be also factored as (1-3x)(1-3x), that is, [-(3x-1)][-(3x-1)] ... so, the "answer in the book" should also give (1-3x), if it gave (3x-1). ]

 

[pooface]

yes magnitude is what i was only indicating at. I didnt read over the other solutions posted. yes when dealing with radicals you can introduce extraneous roots.

I just wanted the fellow to see the 3x-1 but noting that it is a magnitude is worthy.
yes real answer is +-(3x-1).
Thank you Robphy.