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#1 2005-07-22 01:44:01

samsoo
Guest

Roots and Radicals

Hi all.

I'm new here and need some help with this maths problem.

sqrroot.JPG

Here's a pic of the question and the answer and I've tried doing a method but my head can't get round this.

Any help appreciated.

Thanks in advance.

#2 2005-07-22 03:04:03

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Roots and Radicals

I tried working backwards by squaring the answer, and found that you actually want to find to square root of:

3/2(x-1)+√(2x²-7x-4)

I wouldn't have any idea how to solve it if you hadn't already given us the answer though.
Squaring is easy, you can just follow a set of rules to get the answer every time.
Isn't it weird how the inverse can be so much more horrible?


Why did the vector cross the road?
It wanted to be normal.

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#3 2005-07-22 05:43:11

samsoo
Guest

Re: Roots and Radicals

Hi thx for the reply.

You're right, it does work if there's a plus sign between 3/2(x-1) and √(2x²-7x-4). I checked the question and there's no plus sign, so I guess it could be a typo error.

I was given a typed up booklet just to familiarise and practise on, I hope there's no more other typos or I'll just get stressed up for doing wrong questions.

#4 2005-07-22 05:48:50

samsoo
Guest

Re: Roots and Radicals

One more thing, is this a typo error agian or correct answer?

simp.JPG

#5 2005-07-22 06:06:04

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Roots and Radicals

√a x √b= √(a x b), so 4√x³y x 4√xy² = 4√(x³y x xy²)

Multiply the terms: 4√((x^4)y³)
4√x^4=x, so taking x out of the 4th root gives x(4√y³)

Your answer should have the 4th root of y³ instead of the square root.
So, yes, minor typo error.


Why did the vector cross the road?
It wanted to be normal.

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#6 2005-07-22 09:50:20

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,555

Re: Roots and Radicals

Another way to look at it is to think that a square root is also x^½ (x raised to the half power)

For example √x² = x is is the same as saying (x²)^½ = x^(2/2) = x^1 = x

So, 4√x³y x 4√xy² = x^(3/4) × y^(1/4) × x^(1/4) × y^(2/4)

Combining terms: x^(4/4) × y^(3/4)

Simplifying: x × y^(3/4)  = x 4√y³  (x times fourth root of y cubed)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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