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Maths - subbing back in

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  • 05-06-2008 6:46pm
    #1
    guX
    Closed Accounts Posts: 96 ✭✭


    Hey, I was just wondering, is it true that in algebra-type questions you lose 3 marks if you don't sub your solution back into the original equation for verification? Even if your answer is right?


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    #2
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    If the question says Find x, and you find x, they're gonna give you full marks.


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    #3
    Fuascailt
    Closed Accounts Posts: 268 ✭✭Fuascailt


    Well in surd equations, if there was an extraneous root, you'd have to check. Dont see why you'd have to otherwise


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    #4
    NanoZoom555
    Closed Accounts Posts: 21 ✭NanoZoom555


    only for inequalities if you had to square both sides to solve the equation,or for log equations


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    #5
    guX
    Closed Accounts Posts: 96 ✭✭guX


    Fuascailt wrote: »
    Well in surd equations, if there was an extraneous root, you'd have to check. Dont see why you'd have to otherwise

    So does that mean that if we're given a surd equation we ignore that the square root or something could be plus-or-minus and just take the plus value?


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    #6
    minty16
    Registered Users Posts: 561 ✭✭✭minty16


    Get the answer, get the marks. That subbing back in is pointless but useful if you think you might have it wrong.
    Apart from differentiation with respect to first principles where all steps must be clearly written, get the answer and you're alright mate.


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    #7
    22diamonds
    Registered Users Posts: 96 ✭✭22diamonds


    You should check answer in surd questions and log questions. everything else should be ok.


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    #8
    MathsManiac
    Registered Users Posts: 1,595 ✭✭✭MathsManiac


    And if you want to know why you need to check it when there are logs or square roots knocking around, see the marking scheme for 2005 P1 Q5(a). You appear to get two roots, but when you check, only one of them works. A penalty was applied (just a slip on that occasion) for failing to properly deal with this.


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    #9
    BMH
    Closed Accounts Posts: 804 ✭✭✭BMH


    MathsManiac wrote: »
    And if you want to know why you need to check it when there are logs or square roots knocking around, see the marking scheme for 2005 P1 Q5(a). You appear to get two roots, but when you check, only one of them works. A penalty was applied (just a slip on that occasion) for failing to properly deal with this.

    While you're knocking about MathsManiac...
    x^2 = x+x+x+x+...x times
    d(x^2)/dx=2x but d(x+x+x+x...x times)/dx=(1+1+1+1...x times)=x
    2x=!x...


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    #10
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    BMH wrote: »
    While you're knocking about MathsManiac...
    x^2 = x+x+x+x+...x times
    d(x^2)/dx=2x but d(x+x+x+x...x times)/dx=(1+1+1+1...x times)=x
    2x=!x...

    x isn't a constant so you'd need to use the product rule, which will get you 2x on the RHS.


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    #11
    BMH
    Closed Accounts Posts: 804 ✭✭✭BMH


    Cokehead Mother wrote: »
    x isn't a constant so you'd need to use the product rule, which will get you 2x on the RHS.

    But couldn't the RHS be split in d(x)/dx+d(x)/dx+d(x)/dx+d(x)/dx...x times?


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    #12
    Cokehead Mother
    Closed Accounts Posts: 773 ✭✭✭Cokehead Mother


    BMH wrote: »
    But couldn't the RHS be split in d(x)/dx+d(x)/dx+d(x)/dx+d(x)/dx...x times?

    nope.

    Take xsinx. you can't split it up into sinx + sinx + sinx ... x times and differentiate to get xcosx.

    What you're doing is saying that there is a fixed number of d(x)/dx but there isn't.


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    #13
    BMH
    Closed Accounts Posts: 804 ✭✭✭BMH


    Cokehead Mother wrote: »
    nope.

    Take xsinx. you can't split it up into sinx + sinx + sinx ... x times and differentiate to get xcosx.

    What you're doing is saying that there is a fixed number of d(x)/dx but there isn't.

    Ah, of course, thanks.


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