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Topic review (newest first)
- MathsIsFun
- 2005-07-23 07:50:20
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)
- mathsyperson
- 2005-07-23 04:06:04
√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.
- samsoo
- 2005-07-23 03:48:50
One more thing, is this a typo error agian or correct answer?
- samsoo
- 2005-07-23 03:43:11
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.
- mathsyperson
- 2005-07-23 01:04:03
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?
- samsoo
- 2005-07-22 23:44:01
Hi all.
I'm new here and need some help with this maths problem.
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.