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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 $\displaystyle \frac{(2x - 5)(x + 1)}{x} - 1$ or $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$?

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

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

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

    So $\displaystyle x \neq 0$ in the first and $\displaystyle x \neq 1$ in the second.
    is the second one $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$
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    Quote Originally Posted by may View Post
    is the second one $\displaystyle \frac{(2x - 5)(x + 1)}{x - 1}$
    Your statement is still false, as $\displaystyle x$ can not take on the value of $\displaystyle 1$.
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    Quote Originally Posted by Prove It View Post
    Your statement is still false, as $\displaystyle x$ can not take on the value of $\displaystyle 1$.
    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 $\displaystyle f(x) = \frac{(2x - 5)(x + 1)}{x - 1}$. If we can establish a full bijection between f$\displaystyle (x)$ and $\displaystyle x$, then $\displaystyle f(x)$ must take all real values. Let's see, say $\displaystyle f(x) = y$ :

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

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

    $\displaystyle y^2 - 2y + 49$

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

    Therefore the equation :

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

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

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

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