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Core 1 - Questions involving "Square root"

Question 1: Express 9−129^\frac{-1}{2} as a fraction

9−129^\frac{-1}{2} = 19\frac{1}{\sqrt 9} = 1±3\frac{1}{\pm3}

But why is the mark scheme giving the answer as 13\frac{1}{3}? Am I just being being pedantic or am I mathematically incorrect and would be penalised for this?

Question 2: Find the gradient of the curve y=2x+6xy=2x+\frac{6}{\sqrt x} at the point where x=4x=4

y=2x+6xy=2x+\frac{6}{\sqrt x}
y=2x+6x−12y=2x+6x^\frac{-1}{2}

⇒dydx=2−3x−32=2−3x3\Rightarrow \frac{dy}{dx}=2-3x^\frac{-3}{2} = 2-\frac{3}{\sqrt x^3}

Let x=4x=4

⇒2−343=2−364=2−3±8=2\Rightarrow 2-\frac{3}{\sqrt 4^3} = 2-\frac{3}{\sqrt 64} = 2-\frac{3}{\pm8} = 2 ±\pm 38\frac{3}{8}

Therefore gradient at 44 is 138\frac{13}{8} and 198\frac{19}{8}

But why is it on the mark scheme they are only looking for 138\frac{13}{8}?

I'm so confused now when I'm supposed to take the ∣x∣|\sqrt{x}| or ±\pm x\sqrt{x}. I just don't want to get penalised for putting down two answers when there should only be one. :mad:

Thank you very much, I'm looking forward to any advice :smile: Good luck with your exams everyone!

Solution to the Square root issue
So, as far as I understand from these comments, if you've been asked to find the x\sqrt x then you leave it as a positive answer (since they would have asked you ±\pm x\sqrt x if they wanted both values). But when you are trying to solve something such as x2=2x^2=2 then you're answer would be x=±x=\pm 2\sqrt 2. So basically, always take ±\pm unless you've been given the square root initially (with the addition/subtraction symbol before it).

Another way of thinking about it: If the x\sqrt x was always ±\pm then it would make putting b2±b2−4acb^2 \pm \sqrt {b^2-4ac} obsolete since they could have just put −b2−4ac- \sqrt {b^2-4ac} or even ++.
(edited 11 years ago)

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Reply 1
Avatar for raheem94
raheem94
13
Original post by BobGreggary
Question 1: Express 9−129^\frac{-1}{2} as a fraction

9−129^\frac{-1}{2} = 19\frac{1}{\sqrt 9} = 1±3\frac{1}{\pm3}

But why is the mark scheme giving the answer as 13\frac{1}{3}? Am I just being being pedantic or am I mathematically incorrect and would be penalised for this?


The mark scheme answer is correct.

If you have, x2=2 x^2 = 2 , then the answer is, x=±2 x= \pm \sqrt2

But when you have a square root, then you don't need to add plus minus. e.g. 16=4  and  −16=−4 \sqrt{16} = 4 \ \ and \ \ -\sqrt{16} = -4
Original post by raheem94
The mark scheme answer is correct.

If you have, x2=2 x^2 = 2 , then the answer is, x=±2 x= \pm \sqrt2

But when you have a square root, then you don't need to add plus minus. e.g. 16=4  and  −16=−4 \sqrt{16} = 4 \ \ and \ \ -\sqrt{16} = -4


I didn't know this! And I'm doing C4! I've always been pedantic about two answers etc. I suppose it makes perfect sense that a function must only map to one value.
Reply 3
Avatar for elldeegee
elldeegee
19
but if x2=25 x^2 = 25 then the answer is ±25=±5 \pm \sqrt{25} = \pm 5
Reply 4
Avatar for raheem94
raheem94
13
Original post by BobGreggary

Question 2: Find the gradient of the curve y=2x+6xy=2x+\frac{6}{\sqrt x} at the point where x=4x=4

y=2x+6xy=2x+\frac{6}{\sqrt x}
y=2x+6x−12y=2x+6x^\frac{-1}{2}

⇒dydx=2−3x−32=2−3x3\Rightarrow \frac{dy}{dx}=2-3x^\frac{-3}{2} = 2-\frac{3}{\sqrt x^3}

Let x=4x=4

⇒2−343=2−364=2−3±8=2\Rightarrow 2-\frac{3}{\sqrt 4^3} = 2-\frac{3}{\sqrt 64} = 2-\frac{3}{\pm8} = 2 ±\pm 38\frac{3}{8}

Therefore gradient at 44 is 138\frac{13}{8} or 198\frac{19}{8}

But why is it on the mark scheme they are only looking for 138\frac{13}{8}?

I'm so confused now when I'm supposed to take the ∣x∣|\sqrt{x}| or ±\pm x\sqrt{x}. I just don't want to get penalised for putting down two answers when there should only be one. :mad:

Thank you very much, I'm looking forward to any advice :smile: Good luck with your exams everyone!



2−364=2−38=138 \displaystyle 2-\frac{3}{\sqrt 64} = 2 - \frac38 = \frac{13}8
Reply 5
Avatar for TenOfThem
TenOfThem
18
Original post by elldeegee
but if x2=25 x^2 = 25 then the answer is ±25=±5 \pm \sqrt{25} = \pm 5


Yes

But the reason that you need to put the ±\pm is because the x\sqrt{x} is only + on its own
Reply 6
Avatar for raheem94
raheem94
13
Original post by Junaid96
I didn't know this! And I'm doing C4! I've always been pedantic about two answers etc. I suppose it makes perfect sense that a function must only map to one value.


Are you trolling?

The OP was making this mistake, so i tried to point it out. What's wrong in it?
Reply 7
Avatar for elldeegee
elldeegee
19
Original post by TenOfThem
Yes

But the reason that you need to put the ±\pm is because the x\sqrt{x} is only + on its own


so if the square root is included in a function then it is no longer ± \pm ?
Reply 8
Avatar for TenOfThem
TenOfThem
18
Original post by elldeegee
so if the square root is included in a function then it is no longer ± \pm ?


x\sqrt{x} needs the ±\pm if you want to use both values

4=2\sqrt{4} = 2 but the solution to x2 = 4 is ±2\pm2 because the solution is ±4\pm\sqrt{4}
(edited 11 years ago)
Reply 9
Avatar for f1mad
f1mad
13
Original post by BobGreggary
Question 1: Express 9−129^\frac{-1}{2} as a fraction

9−129^\frac{-1}{2} = 19\frac{1}{\sqrt 9} = 1±3\frac{1}{\pm3}

But why is the mark scheme giving the answer as 13\frac{1}{3}? Am I just being being pedantic or am I mathematically incorrect and would be penalised for this?


Because the square root of a number is taken to be the positive square root.

The +- situation occurs when solving an equation.
Original post by elldeegee
so if the square root is included in a function then it is no longer ± \pm ?


The
Unparseable latex formula:

\sqrt

symbol and the index 12\frac{1}{2} are defined as being positive.

If you want both roots you need a ±\pm sign (e.g. the quadratic formula).
Reply 11
Avatar for BobGreggary
BobGreggary
OP
2
Original post by TenOfThem
x\sqrt{x} needs the ±\pm if you want to use both values

4=2\sqrt{4} = 2 but the solution to x2 = 4 is ±2\pm2 because the solution is ±4\pm\sqrt{4}


So technically, when the teachers were all being cocky in primary school saying
the 4=\sqrt{4}= isn't 22 but ±2\pm 2, they were wrong :P
Original post by BobGreggary
So technically, when the teachers were all being cocky in primary school saying
the 4=\sqrt{4}= isn't 22 but ±2\pm 2, they were wrong :P


Depends what they were saying really :smile:
Reply 13
Avatar for raheem94
raheem94
13
Original post by BobGreggary
So technically, when the teachers were all being cocky in primary school saying
the 4=\sqrt{4}= isn't 22 but ±2\pm 2, they were wrong :P


The teacher would probably be saying that x2=4 x^2 = 4 , gives, x=±2 x =\pm 2
Original post by BobGreggary
So technically, when the teachers were all being cocky in primary school saying
the 4=\sqrt{4}= isn't 22 but ±2\pm 2, they were wrong :P


Either

a) you didn't understand what they were saying

or

b) they didn't understand what they were saying
Original post by Mr M
Either

a) you didn't understand what they were saying

or

b) they didn't understand what they were saying

or

c) they were just trying to ensure an understanding that (-2) squared is also 4
NB

If the same Primary Teacher told you that you can only take a little number from a big number then .... believe nothing they ever told you
(edited 11 years ago)
Original post by TenOfThem
NB

If the same Primary Teacher told you that you cannot take a little number from a big number then .... believe nothing they ever told you


Not sure who is more confused. You or the Primary Teacher.
Original post by raheem94
Are you trolling?

The OP was making this mistake, so i tried to point it out. What's wrong in it?


I'm not trolling :smile:

I genuinely would have put two solutions and would have been baffled if I'd seen the mark scheme.
Reply 19
Avatar for raheem94
raheem94
13
Original post by Junaid96
I'm not trolling :smile:

I genuinely would have put two solutions and would have been baffled if I'd seen the mark scheme.


I am surprised, i thought most students will know this.

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