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    Solve: q^2 - 4q + 4 > 0, Answer q is not equal to 2

    My working:
    (q-2)^2 - 4 + 4 > 0
    (q-2)^2 > 0
    (q-2) > 0 -> I've square rooted at this point
    q > 2

    I'm only getting one solution, so I can't sketch it, why is it not equal to two? Is my answer correct?
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    (Original post by Magenta96)
    Solve: q^2 - 4q + 4 > 0, Answer q is not equal to 2

    My working:
    (q-2)^2 - 4 + 4 > 0
    (q-2)^2 > 0
    (q-2) > 0 -> I've square rooted at this point
    q > 2

    I'm only getting one solution, so I can't sketch it, why is it not equal to two? Is my answer correct?
    You can sketch it. Show the quadratic with the minimum just touching the x axis ...
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    (Original post by Mr M)
    You can sketch it. Show the quadratic with the minimum just touching the x axis ...
    Ok, so should I just draw a graph with the minimum at (2,0), and say the one solution is at q > 2?
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    (Original post by Magenta96)
    Ok, so should I just draw a graph with the minimum at (2,0), and say the one solution is at q > 2?
    I don't like your squaring rooting step at all.

    Draw y=(x-2)^2

    Write down the one value where y = 0.

    Note y is never less than 0.

    Knowing this when is y > 0 ?
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    (Original post by Magenta96)
    Solve: q^2 - 4q + 4 > 0, Answer q is not equal to 2

    My working:
    (q-2)^2 - 4 + 4 > 0
    (q-2)^2 > 0
    (q-2) > 0 -> I've square rooted at this point
    q > 2

    I'm only getting one solution, so I can't sketch it, why is it not equal to two? Is my answer correct?
    For the square root
    \displaystyle \sqrt{(q-2)^2}=|q-2|
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    (Original post by Mr M)
    I don't like your squaring rooting step at all.

    Draw y=(x-2)^2

    Write down the one value where y = 0.

    Note y is never less than 0.

    Knowing this when is y > 0 ?
    y is zero at x = 2, and y is > 0 at x < 2 and x > 2
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    (Original post by Magenta96)
    y is zero at x = 2, and y is > 0 at x < 2 and x > 2
    Yes.

    You can write x \neq 2 as an alternative to x&lt;2 and x&gt;2 .
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    (Original post by Mr M)
    Yes.

    You can write x \neq 2 as an alternative to x&lt;2 and x&gt;2 .
    Thank you!
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    (Original post by Magenta96)
    Thank you!
    You are welcome.
 
 
 
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