Core 1 - Questions involving "Square root"

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

    9^\frac{-1}{2} = \frac{1}{\sqrt 9} = \frac{1}{\pm3}

    But why is the mark scheme giving the answer as \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+\frac{6}{\sqrt x} at the point where x=4

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

    \Rightarrow \frac{dy}{dx}=2-3x^\frac{-3}{2} = 2-\frac{3}{\sqrt x^3}

    Let x=4

    \Rightarrow 2-\frac{3}{\sqrt 4^3} = 2-\frac{3}{\sqrt 64} = 2-\frac{3}{\pm8} = 2 \pm \frac{3}{8}

    Therefore gradient at 4 is \frac{13}{8} and \frac{19}{8}

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

    I'm so confused now when I'm supposed to take the |\sqrt{x}| or \pm \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 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 \sqrt x then you leave it as a positive answer (since they would have asked you \pm \sqrt x if they wanted both values). But when you are trying to solve something such as x^2=2 then you're answer would be x=\pm \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 \sqrt x was always \pm then it would make putting b^2 \pm \sqrt {b^2-4ac} obsolete since they could have just put - \sqrt {b^2-4ac} or even +.
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    (Original post by BobGreggary)
    Question 1: Express 9^\frac{-1}{2} as a fraction

    9^\frac{-1}{2} = \frac{1}{\sqrt 9} = \frac{1}{\pm3}

    But why is the mark scheme giving the answer as \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,  x^2 = 2 , then the answer is,  x= \pm \sqrt2

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

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

    But when you have a square root, then you don't need to add plus minus. e.g.  \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.
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    but if  x^2 = 25 then the answer is  \pm \sqrt{25} = \pm 5
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    (Original post by BobGreggary)
    Question 2: Find the gradient of the curve y=2x+\frac{6}{\sqrt x} at the point where x=4

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

    \Rightarrow \frac{dy}{dx}=2-3x^\frac{-3}{2} = 2-\frac{3}{\sqrt x^3}

    Let x=4

    \Rightarrow 2-\frac{3}{\sqrt 4^3} = 2-\frac{3}{\sqrt 64} = 2-\frac{3}{\pm8} = 2 \pm \frac{3}{8}

    Therefore gradient at 4 is \frac{13}{8} or \frac{19}{8}

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

    I'm so confused now when I'm supposed to take the |\sqrt{x}| or \pm \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 Good luck with your exams everyone!

     \displaystyle 2-\frac{3}{\sqrt 64} = 2 - \frac38 = \frac{13}8
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    (Original post by elldeegee)
    but if  x^2 = 25 then the answer is  \pm \sqrt{25} = \pm 5
    Yes

    But the reason that you need to put the \pm is because the \sqrt{x} is only + on its own
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    (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?
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    (Original post by TenOfThem)
    Yes

    But the reason that you need to put the \pm is because the \sqrt{x} is only + on its own
    so if the square root is included in a function then it is no longer  \pm ?
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    (Original post by elldeegee)
    so if the square root is included in a function then it is no longer  \pm ?
    \sqrt{x} needs the \pm if you want to use both values

    \sqrt{4} = 2 but the solution to x2 = 4 is \pm2 because the solution is \pm\sqrt{4}
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    (Original post by BobGreggary)
    Question 1: Express 9^\frac{-1}{2} as a fraction

    9^\frac{-1}{2} = \frac{1}{\sqrt 9} = \frac{1}{\pm3}

    But why is the mark scheme giving the answer as \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.
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    (Original post by elldeegee)
    so if the square root is included in a function then it is no longer  \pm ?
    The \sqrt symbol and the index \frac{1}{2} are defined as being positive.

    If you want both roots you need a \pm sign (e.g. the quadratic formula).
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    (Original post by TenOfThem)
    \sqrt{x} needs the \pm if you want to use both values

    \sqrt{4} = 2 but the solution to x2 = 4 is \pm2 because the solution is \pm\sqrt{4}
    So technically, when the teachers were all being cocky in primary school saying
    the \sqrt{4}= isn't 2 but \pm 2, they were wrong :P
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    (Original post by BobGreggary)
    So technically, when the teachers were all being cocky in primary school saying
    the \sqrt{4}= isn't 2 but \pm 2, they were wrong :P
    Depends what they were saying really
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    (Original post by BobGreggary)
    So technically, when the teachers were all being cocky in primary school saying
    the \sqrt{4}= isn't 2 but \pm 2, they were wrong :P
    The teacher would probably be saying that  x^2 = 4 , gives,  x =\pm 2
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    (Original post by BobGreggary)
    So technically, when the teachers were all being cocky in primary school saying
    the \sqrt{4}= isn't 2 but \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
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    (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
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    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
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    (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.
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    (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

    I genuinely would have put two solutions and would have been baffled if I'd seen the mark scheme.
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    (Original post by Junaid96)
    I'm not trolling

    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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