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quadratic equation question

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    question:
    if  a > 0 , prove that the quadratic expression  ax^2 + bx + c is positive for all real values of x when  b^2 < 4ac .

    my answer:
    let  f(x) = ax^2 + bx + c
      = a\left [ x^2 + \frac{b}{a}x + \frac{c}{a} \right ]

     =a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}
    if f(x) >0 for all real values of x, it has no real roots

    \therefore b^2 - 4ac  wil produce a negative number let call it -g = d

      \therefore \sqrt{ -g}  \rightarrow \mathbb{C}
    since  b^2 - 4ac = -g

     \therefore - g < 0  \rightarrow b^2 - 4ac < 0 then  b^2 < 4ac
     \therefore when
     b^2 < 4ac , f(x) > 0 for all real values of x



    Please I want to know if my proof correctly answers the question please tell where i went wrong
    thank you
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    (Original post by bigmansouf)
    question:
    if  a > 0 , prove that the quadratic expression  ax^2 + bx + c is positive for all real values of x when  b^2 < 4ac .

    my answer:
    let  f(x) = ax^2 + bx + c
      = a\left [ x^2 + \frac{b}{a}x + \frac{c}{a} \right ]

     =a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}
    if f(x) >0 for all real values of x, it has no real roots

    \therefore b^2 - 4ac  wil produce a negative number let call it -g = d

      \therefore \sqrt{ -g}  \rightarrow \mathbb{C}
    since  b^2 - 4ac = -g

     \therefore - g < 0  \rightarrow b^2 - 4ac < 0 then  b^2 < 4ac
     \therefore [\tex] when [tex] b^2 < 4ac f(x) > 0 for all real values of x



    Please I want to know if my proof correctly answers the question please tell where i went wrong
    thank you
    Firstly, you should have more working when you complete the square.

    You are asked to prove that if b^2<4ac then f(x) >0. What you have attempted to do instead is prove that if f(x) >0 then b^2<4ac and then use this to prove f(x) >0.

    But you don't need the first part since you're told already that b^2<4ac. The major problem is that knowhere in your working have you actually proved why b^2<4ac implies that f(x)>0.

    You could talk about the discriminant/no real roots to show this but I assume from your working that you need to prove it by completing the square? You correctly completed the squre but didn't use it for anything.

    Think about why this

    \displaystyle  =a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}

    must always be positive.


    One more thing:

    \sqrt{ -g}  \rightarrow \mathbb{C}

    This doesn't mean much. I assume you're trying to say that \sqrt{ -g} is a complex number but the set of complex numbers includes the reals and the imaginary numbers. What you mean to say is that \sqrt{ -g} is imaginary. There is a mathematical way to write this but you may as well just write it in words and I recommend this unless you're confident with using the formal notation.
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    (Original post by notnek)
    Firstly, you should have more working when you complete the square.

    You are asked to prove that if b^2<4ac then f(x) >0. What you have attempted to do instead is prove that if f(x) >0 then b^2<4ac and then use this to prove f(x) >0.

    But you don't need the first part since you're told already that b^2<4ac. The major problem is that knowhere in your working have you actually proved why b^2<4ac implies that f(x)>0.

    You could talk about the discriminant/no real roots to show this but I assume from your working that you need to prove it by completing the square? You correctly completed the squre but didn't use it for anything.

    Think about why this

    \displaystyle  =a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}

    must always be positive.


    One more thing:

    \sqrt{ -g}  \rightarrow \mathbb{C}

    This doesn't mean much. I assume you're trying to say that \sqrt{ -g} is a complex number but the set of complex numbers includes the reals and the imaginary numbers. What you mean to say is that \sqrt{ -g} is imaginary. There is a mathematical way to write this but you may as well just write it in words and I recommend this unless you're confident with using the formal notation.
    thank you very much for helping me
    here goes my second attempt

    let  f(x) = ax^2 + bx + c
      = a\left [ x^2 + \frac{b}{a}x + \frac{c}{a} \right ]
    = a\left [ x^2 +\frac{b}{a} +\frac{c}{a}\right ]
    = a\left [\left ( x+ \frac{b}{2a} \right )^2 + \frac{c}{a}  - \left ( \frac{b}{2a} \right )^2\right ]
     =a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}


    since  b^2 < 4ac \rightarrow  b^2-4ac<0
    since  a> 0 ,  f(x) >0 has a least value of  \frac{4ac-b^2}{4a} when  x = - \frac{b}{2a}

    since b^2 < 4ac   \rightarrow   \frac{4ac- b^2}{4a} =  \left \{ f(x) \epsilon  \mathbb{R} : f(x) > 0 \right \}
    if   \frac{4ac- b^2}{4a}  is the least values, taking the input values of x that are symmetrical about  x = - \frac{b}{2a} i.e  x = - \frac{b}{2a}\pm k gives  f\left (  - \frac{b}{2a} +k \right ) = f\left (  - \frac{b}{2a} -k \right ) = ak^2 + \frac{4ac-b^2}{4a}

    since the least value  \frac{4ac-b^2}{4a} > 0
     \therefore any inputs of x that are symmetrical will also give output of the same  f(x) where  \left \{ f(x)\epsilon \mathbb{R}: f(x)> 0\cup f(x) > \frac{4ac-b^2}{4a} \right \}

    ( f(x) for points symmetrical about x = -b/2a will be > 0 and > (4ac-b^2) / (4a)
    explaining incase you dont understand

     \therefore when  b^2<4ac ,  f(x) > 0

    Please i understand you said to stick to using words i want to do better and writing mathematical notation was advised by my teacher . if i am wrong with the notation please show me the correct way. I know that the last point which i was saying ( f(x) for points symmetrical about x = -b/2a will be > 0 and > (4ac-b^2) / (4a)
    I want to improve my mathematical notation as i am very slow in writing

    thank you
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    (Original post by bigmansouf)
    thank you very much for helping me
    here goes my second attempt
    I'm going to try explaining this to you because I don't feel like subtle hints are working very welly. (sorry for butting in, notnek!)

    You're over complicating it. From \displaystlye f(x) = a \underbrace{\left (  x+\frac{b}{2a}\right )^2}_{\geq 0}  + \displaystyle \frac{4ac-b^2}{4a}

    you know that

    (i) the squared part is always greater than or equal to 0, because that's a basic property of a square.

    (ii) you're multiplying the square by a, which you know is >0. So a\left(  x+\frac{b}{2a}\right )^2 \geq 0 .

    Now, it would be really nice if we could prove that \frac{4ac - b^2}{4a} was positive, in which case, we would know that f(x) was a sum of two positive terms and hence would be positive itself.

    But, how to do that... well, for a fraction to be positive, we want both its numerator and denominator to be positive. You can see trivially that the denominator is always positive here (because you're given that a > 0, so 4a must also be > 0).

    You're reduced the entire problem to proving that the numerator is also positive. So, how can we prove that 4ac - b^2 > 0? Oh... what a coincidence! This is precisely what we were assuming!

    In writing the actual proof up, it would go something like this:

    f(x) = a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}, since a > 0 the first term is non-negative. Since b^2 < 4ac we can then conclude that 4ac - b^2 > 0 and so \frac{4ac-b^2}{a} > 0 also. So f(x) is the sum of two positive terms and is hence positive itself.

    NB: When I say "two positive terms" I really mean "sum of one non-negative term and positive term", but the latter is a bit of a mouthful.... and the distinction hardly matters there anyway, I've explicitly used non-negative where the distinction does matter.
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    (Original post by Zacken)
    I'm going to try explaining this to you because I don't feel like subtle hints are working very welly. (sorry for butting in, notnek!)

    You're over complicating it. From \displaystlye f(x) = a \underbrace{\left (  x+\frac{b}{2a}\right )^2}_{\geq 0}  + \displaystyle \frac{4ac-b^2}{4a}

    you know that

    (i) the squared part is always greater than or equal to 0, because that's a basic property of a square.

    (ii) you're multiplying the square by a, which you know is >0. So \a\left (  x+\frac{b}{2a}\right )^2 \geq 0.

    Now, it would be really nice if we could prove that \frac{4ac - b^2}{4a} was also positive, in which case, we would know that f(x) was a sum of two positive terms and hence would be positive itself.

    But, how to do that... well, for a fraction to be positive, we want both its numerator and denominator to be positive. You can see trivially that the denominator is always positive here (because you're given that a > 0, so 4a must also be > 0).

    You're reduced the entire problem to proving that the numerator is also positive. So, how can we prove that 4ac - b^2 > 0? Oh... what a coincidence! This is precisely what we were assuming!

    In writing the actual proof up, it would go something like this:

    f(x) = a\left (  x+\frac{b}{2a}\right )^2  + \frac{4ac-b^2}{4a}, since a > 0 the first term is non-negative. Since b^2 < 4ac we can then conclude that 4ac - b^2 > 0 and so \frac{4ac-b^2}{a} > 0 also. So f(x) is the sum of two positive terms and is hence positive itself.
    thank you woow i dont know what to say but thanks

    can i conclude from what you said that if a quadratic function is a sum of two squares therefore f(x) > 0 is positive for all real values of x and thus will have no real roots

    I am just trying to make note on this question
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    (Original post by bigmansouf)
    thank you very much for helping me
    here goes my second attempt
    I think you are over-complicating a quite simple and straight-forward proof. It doesn't flow nicely at all past the completed square form and your notation goes way out of hand.

    Once you complete the square, you're left with \displaystyle f(x)=a(x-\frac{b}{2a})^2+\frac{4ac-b^2}{4a}

    From here you know that: (x-\frac{b}{2a})^2 \geq 0 (ALWAYS due to squaring)

    One of the conditions you are given is a>0 therefore a(x-\frac{b}{2a})^2 \geq 0

    The other condition is b^2<4ac therefore \frac{4ac-b^2}{4a}>0

    Summing the two terms greater than 0 will obviously give you something greater than 0 therefore f(x)>0

    :qed:

    Why are you talking about the roots???

    EDIT: Nevermind, Zacken beat me to it.
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    (Original post by bigmansouf)
    thank you woow i dont know what to say but thanks
    Notnek deserves it more than I do.

    can i conclude from what you said that if a quadratic function is a sum of two squares therefore f(x) > 0 is positive for all real values of x and thus will have no real roots

    I am just trying to make note on this question
    Indeed! (although this is not the case in this particular question) - here it's "if a function is a sum of two positive things for all real values of x, then the function is positive for all real values of x".

    In another question, where you could have f(x) = (x-a)^2 then you can't quite say that f(x) > 0, the best you can say is that f(x) \geq 0, because y^2 \geq 0, not y > 0.

    However, if you had something like f(x) = (x-a)^2 + (x-b)^2 then you can definitely say that f(x) > 0 and so it has no real roots, since both brackets can't be simultaneously 0 for distinct a,b.

    In fact, you use this technique a lot when you're asked to do stuff like "prove that f(x) = x^2 + 2x + 2 has no real roots". You do: f(x) = x^2 + 2x + 1  +1 = \underbrace{(x+1)^2}_{\geq 0} + 1 so that f(x) \geq 0 + 1 > 0 means that f(x) has no real roots.
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    (Original post by RDKGames)
    ...
    Might be worth clearing up your inequalities as well, to prevent further confusion; it's y^2 \geq 0, not y >0 .
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    (Original post by Zacken)
    Notnek deserves it more than I do.



    Indeed! (although this is not the case in this particular question) - here it's "if a function is a sum of two positive things for all real values of x, then the function is positive for all real values of x".

    In another question, where you could have f(x) = (x-a)^2 + (x-b)^2 then you can't quite say that f(x) > 0, the best you can say is that f(x) \geq 0, because y^2 \geq 0, not y > 0.

    However, if you had something like f(x) = (x-a)^2 + (x-b)^2 + 1 then you can definitely say that f(x) > 0 and so it has no real roots.

    In fact, you use this technique a lot when you're asked to do stuff like "prove that f(x) = x^2 + 2x + 2 has no real roots". You do: f(x) = x^2 + 2x + 1  +1 = \underbrace{(x+1)^2}_{\geq 0} + 1 so that f(x) \geq 0 + 1 > 0 means that f(x) has no real roots.
    thank i gave him a rep so did u and RDK games

    i am going to work harder i feel i got played around by bostock and chandler (making easy stuff look difficult)
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    (Original post by bigmansouf)
    thank i gave him a rep so did u and RDK games

    i am going to work harder i feel i got played around by bostock and chandler (making easy stuff look difficult)
    Good man.

    That's a good idea! You should probably start off with easier resources and then use B&C to supplement your learning instead of learning solely from it - some of the stuff is a bit archaic.

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