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Thread: Why is discriminant taken to be zero when finding the range of y?

  1. #1
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    Why is discriminant taken to be zero when finding the range of y?

    So, I was given a question which went like this

    y=Quadratic/Quadratic, find range of y.

    So, my teacher said that I needed to make a quadratic in x by taking all the terms in one side and then I had to make the discriminant greater than or equal to zero since x is real. But, discriminant is greater than or equal to zero when roots are real and distinct. So, I am unable to relate to that theory of quadratic equation.
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    Re: Why is discriminant taken to be zero when finding the range of y?

    Quote Originally Posted by abhistud View Post
    So, I was given a question which went like this

    y=Quadratic/Quadratic, find range of y.
    I'm not sure what you are talking about. Do you mean a fraction in which both numerator and denominator are quadratic in x? Could you give a specific example?

    So, my teacher said that I needed to make a quadratic in x by taking all the terms in one side and then I had to make the discriminant greater than or equal to zero since x is real. But, discriminant is greater than or equal to zero when roots are real and distinct. So, I am unable to relate to that theory of quadratic equation.
    No, if the discriminant is greater than 0, the roots are real and distinct. When the discriminant is 0 the roots are real but not distinct. There is one real root.
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    Re: Why is discriminant taken to be zero when finding the range of y?

    Yeah!
    Say,
    Find the range of
    $\frac{x^2+2x-2}{x^2+2x+1}$
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    Re: Why is discriminant taken to be zero when finding the range of y?

    What I wanted to say was that the roots may be real and distinct or they might be equal.
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    Re: Why is discriminant taken to be zero when finding the range of y?

    The domain of the function, the values that x can take, are real. This means that the discriminant must be non-negative because a negative discriminant gives complex values of x.
    Last edited by Archie; Apr 4th 2017 at 04:17 AM.
    Thanks from abhistud
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    Re: Why is discriminant taken to be zero when finding the range of y?

    Quote Originally Posted by abhistud View Post
    Yeah!
    Say,
    Find the range of
    $\frac{x^2+2x-2}{x^2+2x+1}$
    This is quite different from the question you first asked. The first thing I would do the indicated division to write this as f(x)= 1- \frac{3}{x^+ 2x+ 1}. To find the range of that first find the range of the denominator. And that has nothing to do with finding a "discriminant". You should be able to immediately see that x^2+ 2x+ 1= (x+ 2)^2. This function is f(x)= 1- \frac{3}{(x+ 1)^2}. That fraction takes on all values except 0 (a fraction is never 0 unless the numerator is 0) so f takes on all values except 1.

    I wouldn't worry about a discriminant
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    Re: Why is discriminant taken to be zero when finding the range of y?

    y=\frac{x^2+2x-2}{x^2+2x+1}

    gives a second degree equation in x

    (y-1)x^2+2(y-1)x+(y+2)=0

    y is in the range if and only if this equation has a real solution

    therefore the discriminant 12-12y>0 so the range is y<1
    Thanks from Archie and abhistud
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    Re: Why is discriminant taken to be zero when finding the range of y?

    Quote Originally Posted by HallsofIvy View Post
    This is quite different from the question you first asked. The first thing I would do the indicated division to write this as f(x)= 1- \frac{3}{x^+ 2x+ 1}. To find the range of that first find the range of the denominator. And that has nothing to do with finding a "discriminant". You should be able to immediately see that x^2+ 2x+ 1= (x+ 2)^2. This function is f(x)= 1- \frac{3}{(x+ 1)^2}. That fraction takes on all values except 0 (a fraction is never 0 unless the numerator is 0) so f takes on all values except 1.

    I wouldn't worry about a discriminant
    HallsOfIvy has had a nightmare with this answet. Most uncharacteristic. He's right about the discriminant being unnecessary, but it is still a perfectly valid way to approach the question. See Idea's post.
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