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  1. #1
    may
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    Exclamation quadratic equation

    Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.
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    Quote Originally Posted by may View Post
    Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.
    Is this \frac{(2x - 5)(x + 1)}{x} - 1 or \frac{(2x - 5)(x + 1)}{x - 1}?

    In either case, your statement is false, as you can't divide by 0.

    So x \neq 0 in the first and x \neq 1 in the second.
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    Quote Originally Posted by Prove It View Post
    Is this \frac{(2x - 5)(x + 1)}{x} - 1 or \frac{(2x - 5)(x + 1)}{x - 1}?

    In either case, your statement is false, as you can't divide by 0.

    So x \neq 0 in the first and x \neq 1 in the second.
    is the second one \frac{(2x - 5)(x + 1)}{x - 1}
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    Quote Originally Posted by may View Post
    is the second one \frac{(2x - 5)(x + 1)}{x - 1}
    Your statement is still false, as x can not take on the value of 1.
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    Quote Originally Posted by Prove It View Post
    Your statement is still false, as x can not take on the value of 1.
    I think the problem is referring to the range of the expression rather than the domain.
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    Quote Originally Posted by may View Post
    Show that if x is real ,the expression [(2x-5)(x+1)]/x-1 can take all the real values.
    I've never tried this kind of question yet, correct me if I'm wrong.

    Let f(x) = \frac{(2x - 5)(x + 1)}{x - 1}. If we can establish a full bijection between f (x) and x, then f(x) must take all real values. Let's see, say f(x) = y :

    y = \frac{(2x - 5)(x + 1)}{x - 1}
    x = \frac{1}{4} \left ( - \sqrt{y^2 - 2y + 49} + y + 3 \right )

    The square root is bothersome, as we need to show that the value under is (hopefully) always positive. So :

    y^2 - 2y + 49

    Discriminant is \Delta = 2^2 - 4 \times 49 = -192 < 0. Since the quadratic coefficient, a, is positive, this quadratic yields only positive values.

    Therefore the equation :

    x = \frac{1}{4} \left ( - \sqrt{y^2 - 2y + 49} + y + 3 \right )

    Does accept all values of y. And therefore, in :

    y = \frac{(2x - 5)(x + 1)}{x - 1}

    y does take all real values. f(x) does take all real values for x \in \mathbb{R}. QED.
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