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Math Help - P(x)=ax^2+bx+c

  1. #1
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    P(x)=ax^2+bx+c

    True or False?
    If a, b, and c denote real numbers and p(x)=ax^2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

    Please show work
    Last edited by yoman360; June 4th 2009 at 08:02 PM.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by yoman360 View Post
    True or False?
    If a, b, and c denote real numbers and p(x)=ax2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

    Please show work
    if

    b^2-4ac>0

    then there exist two different solutions

    if

    b^2-4ac<0

    there is no solutions

    b^2-4ac=0

    there are two same solutions
    for more information see link below
    Quadratic function - Wikipedia, the free encyclopedia
    Last edited by Amer; June 4th 2009 at 08:35 PM.
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  3. #3
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yoman360 View Post
    True or False?
    If a, b, and c denote real numbers and p(x)=ax2+bx+c, x € ℝ, then there exists a real number x0 such that p(x0)=0.

    Please show work
    False.

    Putting this equation in standard form we have

    p(x)=ax^2+bx+c
    =a(x^2+\frac{b}{a}x)+c
    =a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c
    =a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c
    =a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c

    Assume that a>0, we can conclud that this is graph of a parabola that opens upwards and has vetex (-\frac{b}{2a},c-\frac{b^2}{4a}). Again, since the parabola opens upward, p(x_0)=0\Longleftrightarrow{c-\frac{b^2}{4a}\leq0}

    And since there are enumerable instances where this statement does not hold, the answer is false
    Last edited by VonNemo19; June 14th 2009 at 12:59 PM. Reason: Small touch up
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    False.

    Putting this equation in standard form we have

    p(x)=ax^2+bx+c
    =a(x^2+\frac{b}{a}x)+c
    =a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2})+c
    =a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})-\frac{b^2}{4a}+c
    =a(x+\frac{b}{2a})^2-\frac{b^2}{4a}+c

    Assume that a>0, we can conclud that this is graph of a parabola that opens upwards and has vetex (-\frac{b}{2a},c-\frac{b^2}{4a}). Again, since the parabola opens upward, p(x_0)=0\Longleftrightarrow(-\frac{b}{2a},c-\frac{b^2}{4a})\leq0

    And since there are enumerable instances where this statement does not hold, the answer is false
    what if a<0 then it would cross the x-axis thus making the statement true.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by VonNemo19 View Post
    False.

    ...

    Assume that a>0, we can conclude that this is graph of a parabola that opens upwards and has vertex (-\frac{b}{2a},c-\frac{b^2}{4a}). Again, since the parabola opens upward, p(x_0)=0\Longleftrightarrow{\color{red}(-\frac{b}{2a},c-\frac{b^2}{4a})\leq0}
    The part in red doesn't make sense. What I think you're trying to say is that p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0. An ordered pair cannot be less that a number.

    And since there are enumerable instances where this statement does not hold, the answer is false
    But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

    True or False?
    If a,b,c\in\mathbb{R} and p\!\left(x\right)=ax^2+bx+c, x\in\mathbb{R}, then there exists a real number x_0 such that p\!\left(x_0\right)=0.
    This is a true statement under certain cases. Again, it all depends on the value of the discriminant \Delta = b^2-4ac. If

    \Delta > 0, then there are two solutions (so there exists more than one x_0).

    \Delta = 0 , there there is only one solution (thus x_0 exists).

    \Delta < 0 , there there is no real solution. This implies that a x_0 does not exist.

    To summarize, the only time when an x_0 exists is when \Delta\geq 0. Also, the value of x_0 is the real zero(s) of the quadratic equation.

    Consider f\!\left(x\right)=x^2+3x+2. Since 1,3,2\in\mathbb{R}, we search for an x_0 such that f\!\left(x_0\right)=0. So in essence, we solve x_0^2+3x_0+2=0. But x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+  1\right)=0\implies\left(x_0+2\right)=0 or \left(x_0+1\right)=0. So in this case, there exists two possible values for x_0: x_0=-2 and x_0=-1, where both values for x_0 are real numbers.
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    The part in red doesn't make sense. What I think you're trying to say is that p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0. An ordered pair cannot be less that a number.

    But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

    This is a true statement under certain cases. Again, it all depends on the value of the discriminant \Delta = b^2-4ac. If

    \Delta > 0, then there are two solutions (so there exists more than one x_0).

    \Delta = 0 , there there is only one solution (thus x_0 exists).

    \Delta < 0 , there there is no real solution. This implies that a x_0 does not exist.

    To summarize, the only time when an x_0 exists is when \Delta\geq 0. Also, the value of x_0 is the real zero(s) of the quadratic equation.

    Consider f\!\left(x\right)=x^2+3x+2. Since 1,3,2\in\mathbb{R}, we search for an x_0 such that f\!\left(x_0\right)=0. So in essence, we solve x_0^2+3x_0+2=0. But x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+  1\right)=0\implies\left(x_0+2\right)=0 or \left(x_0+1\right)=0. So in this case, there exists two possible values for x_0: x_0=-2 and x_0=-1, where both values for x_0 are real numbers.
    Oh ok I get it now
    can you help me with this one:
    http://www.mathhelpforum.com/math-he...z-1-2-1-a.html
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  7. #7
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    Wouldn't the answer be true? because 0 is a real number and if b and c= 0 then then p(x_0)=0
    p(x)= ax^2 +bx +c
    p(x) = 1x^2+0x+0
    p(x) = x^2
    then there's only one solution of the function.
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  8. #8
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yoman360 View Post
    Wouldn't the answer be true? because 0 is a real number and if b and c= 0 then then p(x_0)=0
    p(x)= ax^2 +bx +c
    p(x) = 1x^2+0x+0
    p(x) = x^2
    then there's only one solution of the function.
    Yes, that is true. Note that you imposed certain conditions on what you let a,b, and c equal. As I said above, the statement is true for certain cases/conditions, and this would be one of them. You just illustrated the case where we only have one value x_0 such that p\!\left(x_0\right)=0 (I illustrated a case where we had more than one value).
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    The part in red doesn't make sense. What I think you're trying to say is that p\!\left(x_0\right)=0\iff c-\frac{b^2}{4a}\leq0. An ordered pair cannot be less that a number.

    But the question isn't asking if this holds for all cases. The statement is an existential statement; one that asks us to find at least one case where the statement is true.

    This is a true statement under certain cases. Again, it all depends on the value of the discriminant \Delta = b^2-4ac. If

    \Delta > 0, then there are two solutions (so there exists more than one x_0).

    \Delta = 0 , there there is only one solution (thus x_0 exists).

    \Delta < 0 , there there is no real solution. This implies that a x_0 does not exist.


    To summarize, the only time when an x_0 exists is when \Delta\geq 0. Also, the value of x_0 is the real zero(s) of the quadratic equation.

    Consider f\!\left(x\right)=x^2+3x+2. Since 1,3,2\in\mathbb{R}, we search for an x_0 such that f\!\left(x_0\right)=0. So in essence, we solve x_0^2+3x_0+2=0. But x_0^2+3x_0+2=0\implies\left(x_0+2\right)\left(x_0+  1\right)=0\implies\left(x_0+2\right)=0 or \left(x_0+1\right)=0. So in this case, there exists two possible values for x_0: x_0=-2 and x_0=-1, where both values for x_0 are real numbers.
    So, you maintain that there exists a zero for every parabola in the form of ax^2+b^2+c?

    You were right about the last statement of my argument, a small error, but the result is still sound. I cahnged it. But by reading the question again, and taking note of the phrase "there exists" , I interpret this to mean that "there must exist" or 'there always exists". And you and I both know that this is a false statement.

    I will attempt to reword the statement. Note that a,b, and c can be any real numbers.

    --For any given parabola, there exists at least one place where it touches the x-axis.--

    Is this ridiculous, or is this ridiculous?



    "Exists is not a predicate."Immanuel Kant.
    Last edited by VonNemo19; June 14th 2009 at 01:03 PM.
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  10. #10
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    The original statement was, essentially that "there exist a real number solution to every quadratic equation". The simplest way to show that is false is to take a= 1, b= 0, c= 1. There is no real number, x, such that x^2+ 1= 0.
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  11. #11
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    Ok it makes sense now since the statement is not always true its false.
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by yoman360 View Post
    Ok it makes sense now since the statement is not always true its false.
    Indubidubly!
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