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    Can someone please tell me the method to finding out the intercepts of quadratic equations such as 6x^2+4x+9=0
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    someone help plz
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    (Original post by madi24)
    Can someone please tell me the method to finding out the intercepts of quadratic equations such as x^2+14x+49=0
    You can factorise the quadratic you mentioned above:

    (x+7)^2=0

    From this you can see that the x intercept would involve a repeated root which is -7
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    thank you
    what should i do to find the y intercept?
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    for y just calculate f(0) so it's 9
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    (Original post by AndreiMiron)
    for y just calculate f(0) so it's 9
    Way easier method, it's the constant term at the end of the quadratic (the one with no x).
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    (Original post by AmmarTa)
    Way easier method, it's the constant term at the end of the quadratic (the one with no x).
    basically that's what f(0) is ... 6* 0^2 + 4*0 +9
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    (Original post by AndreiMiron)
    basically that's what f(0) ...
    Yes but manually calculating f(0) is so much longer if you do the working out / put it in a calculator. Plus looking at the constant term and being able to identify it as the y-intercept is very useful in later maths as you can use it when working with unknown factors in maths at a higher level:
    x^2+ax+5=
    root is -5, find other root
    therefore can be re-written as (x+5)(x+a)
    5a = 5 as 5 * a gives the constant term 5, therefore a = 5/5 = 1
    (x+5)(x+1)=0
    x+1 = 0, x = -1
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    (Original post by AndreiMiron)
    basically that's what f(0) is ... 6* 0^2 + 4*0 +9
    You're still right though, I'm not bashing your method as it's perfectly acceptable - just pointing out an alternative.
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    (Original post by AmmarTa)
    Yes but manually calculating f(0) is so much longer if you do the working out / put it in a calculator.
    No it isn't... it's the exact same thing. Most people can look at 6*0 and immediately see it's 0, I'm not sure how long that takes you to do.
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    (Original post by Zacken)
    No it isn't... it's the exact same thing. Most people can look at 6*0 and immediately see it's 0, I'm not sure how long that takes you to do.
    Yeah that's true , just a longer process than instant identification of the term though. But that's my own personal opinion so I wouldn't take it too seriously.
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    (Original post by AmmarTa)
    Yes but manually calculating f(0) is so much longer if you do the working out / put it in a calculator. Plus looking at the constant term and being able to identify it as the y-intercept is very useful in later maths as you can use it when working with unknown factors in maths at a higher level:
    x^2+ax+5=
    root is -5, find other root
    therefore can be re-written as (x+5)(x+a)
    5a = 5 as 5 * a gives the constant term 5, therefore a = 5/5 = 1
    (x+5)(x+1)=0
    x+1 = 0, x = -1
    even if your polynomial has parameters it still doesn't matter 0*a , a being a complex number is always 0 ... and why would you ever put such an expression in a calculator
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    (Original post by AndreiMiron)
    even if your polynomial has parameters it still doesn't matter 0*a , a being a complex number is always 0 ... and why would you ever put such an expression in a calculator
    a is never stated to be complex in the question?
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    (Original post by AndreiMiron)
    I was pointing the fact that even if it was it wouldn't matter and it obviously applies to real numbers as C includes R
    Oh alright, sorry about that but it'd be interesting to see a polynomial with imaginary coeffiicients! I've never personally come across one.
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    (Original post by AndreiMiron)
    I was pointing the fact that even if it was it wouldn't matter and it obviously applies to real numbers as C includes R
    This is all a bit cringe.

    the fact that when a parameter type is not specified you have to use the largest possible set
    No??? lmao where do you pull this crap from?

    so it would be the set of complex numbers
    There is no largest possible set. It is obvious that for any set A you can get to a strictly larger set (in cardinality) by taking it's power set. So no, the complex numbers are not the largest possible set, there are infinitely many sets that are 'larger'. If you want to restrict to number fields, then the octonions or some other similar object would be your "larger" set.

    tl;dr your post is mostly nonsense
 
 
 
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