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Math Help - Quadratic with 2 variables. Show that it has real roots.

  1. #1
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    Unhappy Quadratic with 2 variables. Show that it has real roots.

    Hi everyone,

    I'm really stumped on this question. Would really appreciate some guidance... "show that 3vx^2 + (4v-6)x + 2v-5 = 0 has real solutions in x if and only if (v-3)(2v-3)<=0" (lesser than or equal to zero).

    Thank you to anyone who can help...

    symphonicity
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  2. #2
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    Re: Quadratic with 2 variables. Show that it has real roots.

    This is only one variable (x), while the other letter is a parameter, so it and any combinations of it are treated as constants.

    Now to have real roots, the discriminant is nonnegative. Can you write an equation for that situation?
    Thanks from topsquark and HallsofIvy
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  3. #3
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    Re: Quadratic with 2 variables. Show that it has real roots.

    For the equation
    ax^2+bx+c=0

    I'm sure you know that the roots are
    x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
    With this equation you can see that the roots are real if b^2-4ac is not negative.
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