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If x is positive, why take the negative root? Watch

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    (Original post by frostyy)
    Thanks! I understand now. I appreciate the help.
    You're welcome, glad to help!
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    (Original post by frostyy)
    Thank you! Just to be clear on this though, I thought that if I have x outside the brackets on one side of the equation, I should divide everything on the other side by it to take it out?

    So x(4-x) = x^2(7-x)
    4 - x = (x^2 / x)((7 - x) / x)

    I understand that I'm wrong now though.
    No. Don't divide by x.

    You have x(4-x) = x^2(7-x). Move it all over to one side. You pick which. I'll go with x(4-x) - x^2(7-x) = 0.

    This then gets \displaystyle x((4-x) - x(7-x)) = 0 by factoring an x out. Simplifying and expanding the brackets inside gets you x(4-x -7x + x^2) = x(x^2 -8x + 4) = 0 as required by the question in part (ii).

    Also, when even though you're dividing by x (which isn't correct) you're doing it incorrectly.

    Do you think that \displaystyle 10 = 5\times 2 means that if you divide by sides by 10 you get \displaystyle \frac{10}{10} = \frac{5}{10} \times \frac{2}{10}?

    That would get you \displaystyle 1 = \frac{1}{2} \times \frac{2}{10} = \frac{1}{10}. And you know that 1 isn't 0.1?

    Instead, if you divide by 10, you divide the entire RHS by 10 only once. So \displaystyle \frac{10}{10} = \frac{5\ times 2}{10} = 1.

    Analogously: When you have \displaystyle x(4-x) = x^2 \times (7-x) if you were to divide by x (which you shouldn't do!) but if you did for the bantz, you should get:

    \displaystyle \frac{x(4-x)}{x} = \frac{x^2}{x} \times (7-x) = x(7-x) and NOT \displaystyle \frac{x(4-x)}{x} = \frac{x^2}{x} \times \frac{7-x}{x}

    Hope that clears up some stuff.

    tl;dr

    1. ax = b + c then dividing by a gets you x = \frac{b+c}{a} = \frac{b}{a} + \frac{c}{a}.

    2. ax = bc then dividing by a gets you a = \frac{bc}{a} \neq \frac{b}{a} \frac{c}{a}.

    Edit: sorry if this comes across as a bit blunt, it's 2:30 a.m and I want to make sure that you nip all these misconceptions before they become bad habits.
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    (Original post by Zacken)
    No. Don't divide by x.

    You have x(4-x) = x^2(7-x). Move it all over to one side. You pick which. I'll go with x(4-x) - x^2(7-x) = 0.

    This then gets \displaystyle x((4-x) - x(7-x)) = 0 by factoring an x out. Simplifying and expanding the brackets inside gets you x(4-x -7x + x^2) = x(x^2 -8x + 4) = 0 as required by the question in part (ii).

    Also, when even though you're dividing by x (which isn't correct) you're doing it incorrectly.

    Do you think that \displaystyle 10 = 5\times 2 means that if you divide by sides by 10 you get \displaystyle \frac{10}{10} = \frac{5}{10} \times \frac{2}{10}?

    That would get you \displaystyle 1 = \frac{1}{2} \times \frac{2}{10} = \frac{1}{10}. And you know that 1 isn't 0.1?

    Instead, if you divide by 10, you divide the entire RHS by 10 only once. So \displaystyle \frac{10}{10} = \frac{5\ times 2}{10} = 1.

    Analogously: When you have \displaystyle x(4-x) = x^2 \times (7-x) if you were to divide by x (which you shouldn't do!) but if you did for the bantz, you should get:

    \displaystyle \frac{x(4-x)}{x} = \frac{x^2}{x} \times (7-x) = x(7-x) and NOT \displaystyle \frac{x(4-x)}{x} = \frac{x^2}{x} \times \frac{7-x}{x}

    Hope that clears up some stuff.

    tl;dr

    1. ax = b + c then dividing by a gets you x = \frac{b+c}{a} = \frac{b}{a} + \frac{c}{a}.

    2. ax = bc then dividing by a gets you a = \frac{bc}{a} \neq \frac{b}{a} \frac{c}{a}.
    That is a long ass explanation and I'm surprised that you're bothered, but nevermind, I owe you a beer or something. Hit me up when you're in the South West :awesome::awesome:
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    (Original post by frostyy)
    That is a long ass explanation and I'm surprised that you're bothered, but nevermind, I owe you a beer or something. Hit me up when you're in the South West :awesome::awesome:
    Haha, I figured it'd be best if I nip your misconceptions before they become bad habits. That happened to me when I was younger... never again. :no:

    If it's too long, don't bother now, but make sure you go over it with a fine tooth comb tomorrow morning and understand precisely what's going on, it'll be worth your while.
 
 
 
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