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  1. #1
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    Make x the subject in each of the following...

    I got this question but I'm not really sure how one part of the solutions came up

    $\displaystyle tx^2+4tx+10=t $

    Make x the subject in each of the following and give the values of t for which real solution(s) to the equation can be found.

    So I complete the square and get up to

    $\displaystyle tx^2+4tx+10=t $

    $\displaystyle \frac {tx^2+4tx+10-t}{t}=0$

    $\displaystyle x^2+4x=\frac{-10}{t} +1 $ I moved $\displaystyle \frac{10}{t}-1 $ to the other side

    $\displaystyle (x+2)^2 = \frac {-10}{t} +5 $

    $\displaystyle (x+2)^2 = \frac {5(-2+t)}{t} $

    So I just take the square root of both sides and subtract -2.

    But the solutions say $\displaystyle x=-2 \pm\frac {\sqrt{5t(-2+t)}}{t} $

    How come my answer is different...
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    $\displaystyle \displaystyle t\,x^2 + 4t\,x + 10 = t$

    $\displaystyle \displaystyle x^2 + 4x + \frac{10}{t} = 1$

    $\displaystyle \displaystyle x^2 + 4x + 2^2 - 2^2 + \frac{10}{t} = 1$

    $\displaystyle \displaystyle \left(x + 2\right)^2 + \frac{10 - 4t}{t} = \frac{t}{t}$

    $\displaystyle \displaystyle \left(x + 2\right)^2 = \frac{5t - 10}{t}$

    $\displaystyle \displaystyle x + 2 = \pm \sqrt{\frac{5t - 10}{t}}$

    $\displaystyle \displaystyle x + 2 = \pm \sqrt{\frac{5t^2 - 10t}{t^2}}$

    $\displaystyle \displaystyle x + 2 = \pm \frac{\sqrt{5t(t - 2)}}{t}$

    $\displaystyle \displaystyle x = 2\pm \frac{\sqrt{5t(t-2)}}{t}$.


    The solution the book gives (which I have posted) involves rationalising the denominator.
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    Ahh alright, if I get a question like this in an exam or something would I have to rationalise every time? Is it wrong if I don't?
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    Quote Originally Posted by jgv115 View Post
    Ahh alright, if I get a question like this in an exam or something would I have to rationalise every time? Is it wrong if I don't?
    Both answers are correct, but it's always best to have a rational denominator.
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