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Hence Show That Blah Blah Has No Real Roots watch

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    I'm looking over a past paper for OCR Additional Maths and one of the questions has this in it.
    (ii) Hence show that the equation f(x) = 0 has no real roots.
    (iv) Hence show that the equation g(x) = 0 has only one real root.
    I'm not sure what it wants you to do; what is ment be a "real root"?
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    A root is where you find a value of x for which the equation = 0

    e.g. If we took x^2 - 1=0
    Then i could factorise that to (x + 1) (x - 1) And i'd know that it had two real roots of x=1 and x=-1

    You need to use the discriminant (b^2 - 4ac)
    If that is less than 0 then there are no real roots
    If it is more than 0 then there are 2 real roots
    If it = 0 then there is 1 real root
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    A real root is basically the number of times the function crosses the x axis.

    If it is a quadratic, by calculating b²-4ac you will see whether it equals, is less or more than 0. If it equals zero, there is one root. If it more than zero, two roots and if it is less than zero no roots.
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    (Original post by nimajneb)
    I'm looking over a past paper for OCR Additional Maths and one of the questions has this in it.


    I'm not sure what it wants you to do; what is ment be a "real root"?
    Well it depends on the nature of the functions f and g - say if they were quadratics you could look at the discriminant

    For the general case you need to show, one way or another, that the curve y = f(x) and the x-axis cross at no point (or in the second case at one point)

    The equations may, of course, have other complex roots - if you have met complex numbers
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    Ok, well the first part gives you "(x+3)²+2" and the lowest value that can give you is 2, which wouldn't cut the x axis; would it be ok writing that in the exam?

    How would I find if it only has one real root (in the seccond case)?
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    (Original post by nimajneb)
    Ok, well the first part gives you "(x+3)²+2" and the lowest value that can give you is 2, which wouldn't cut the x axis; would it be ok writing that in the exam?

    How would I find if it only has one real root (in the seccond case)?
    A quadratic equation ax^2 + bx + c = 0 has

    (i) two distinct real roots when b^2 - 4ac >0

    (ii) one repeated real root when b^2 - 4ac = 0

    (iii) no real roots (they are a conjugate pair of complex numbers) when b^2 - 4ac < 0
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    You would expand (x+3)²+2 = x²+6x+9+2=x²+6x+11
    And then you would find the discriminant: b²-4ac = 36-4*1*11=36-44=-8
    If b²-4ac had equalled 0, it would have one root. However, it doesn't, and in fact it is less than zero, therefore there are no roots.

    It is also valid to say that it has no roots since the lowest y value is 2, as this means that it would not touch the x axis.
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    Hmm, I'm still not quite sure how to go about it.
    You are given that g(x) = x³ + 4x² - x -22
    (iii) Show that g(2) = 0.
    2³ + 4(2²) - 2 -22
    => 8 +16 -24 = 0


    (iv) Hence show that the equation g(x) = 0 has only one real root.
    Would I just say that g(2) = 0 so it just touchs the x axis giving only 1 root?
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    (Original post by nimajneb)
    Hmm, I'm still not quite sure how to go about it.


    Would I just say that g(2) = 0 so it just touchs the x axis giving only 1 root?
    No you need to do more than that. Showing g(2)=0 means it has a root but it might cross and recross at other points

    Factorising the cubic you'll see

    x^3 + 4x^2 - x - 22 = (x-2)(x^2 + 6x + 11)

    and then showing (say using the discriminant again) that the quadratic factor has no real roots
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    (Original post by nimajneb)
    Hmm, I'm still not quite sure how to go about it.


    Would I just say that g(2) = 0 so it just touchs the x axis giving only 1 root?
    Well now that you know that g(2) = 0, then 2 is a root using the factor theorem you get (x-2) is a factor.
    So now using algebraic long division, divide (x³+4x²-x-22) by (x-2)
    when you do this you can factorise getting:

    (x-2)(x²+6x+11)

    Now this equation shows that (x-2) is factor, but (x²+6x+11) cannot be further factored, to prove this you can use the discriminant from quadratic formula and hence state that there is only one factor hence one root (2)

    Ofcourse if you draw a very accurate graph of (x³+4x²-x-22) and show that it only touches x-axis once at 2, then this would also do.
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    Thanks alot.
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    (Original post by Vijay1)
    Ofcourse if you draw a very accurate graph of (x³+4x²-x-22) and show that it only touches x-axis once at 2, then this would also do.
    Yes you could show that the cubic has no turning points or if it has that they are on the same side of the y-axis
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    (Original post by nimajneb)
    I'm looking over a past paper for OCR Additional Maths and one of the questions has this in it.


    I'm not sure what it wants you to do; what is ment be a "real root"?
    if the graph crosses the x-axis teh eqtn has a real root - if it passes above it it is an unreal root - i.e imaginary in the sense that there is no physical crossing
 
 
 
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