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Math Help - Another change the subject question

  1. #1
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    Another change the subject question

    Not sure where to start with this one

    a=2\sqrt{b^2-2} (Make b the subject of the formula)

    How do I begin please? I can't see how to square everything at all this time, not even incorrectly.
    Last edited by Meggomumsie; September 27th 2009 at 09:15 AM. Reason: Forgot to identify the new subject for the formula
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  2. #2
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    Quote Originally Posted by Meggomumsie View Post
    Not sure where to start with this one

    a=2\sqrt{b^2-2} (Make b the subject of the formula)

    How do I begin please? I can't see how to square everything at all this time, not even incorrectly.
    a = 2\sqrt{b^2 - 2}

    a^2 = (2\sqrt{b^2 - 2})^2

    a^2 = 4(b^2 - 2)

    \frac{a^2}{4} = b^2 - 2

    \frac{a^2}{4} + 2 = b^2

    \frac{a^2 + 8}{4} = b^2

    \sqrt{\frac{a^2 + 8}{4}} = b

    b = \frac{\sqrt{a^2 + 8}}{2}.
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  3. #3
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    Quote Originally Posted by Prove It View Post

    b = \textcolor{red}{\pm}\frac{\sqrt{a^2 + 8}}{2}.
    ...
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  4. #4
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    Quote Originally Posted by skeeter View Post
    ...
    Of course...

    I didn't want to give EVERYTHING away
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  5. #5
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    Quote Originally Posted by skeeter View Post
    ...
    b > \sqrt{2}

     a \geq 0 and since a^2 +8 > 0 the negative solution can be discarded

    unless the OP is including the complex numbers
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  6. #6
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    Thanks for your answer I understand what you have done, but I don't understand how we get from

    a^2=(2\sqrt{b^2-2})^2

    to

    a^2=4(b^2-2)

    Can you explain please?
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  7. #7
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    Quote Originally Posted by e^(i*pi) View Post
    b > \sqrt{2}
    correction ...

    b^2 - 2 \ge 0

    b \ge \sqrt{2} or b \le -\sqrt{2}

    hence the need for the \pm
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  8. #8
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    Quote Originally Posted by Meggomumsie View Post
    Thanks for your answer I understand what you have done, but I don't understand how we get from

    a^2=(2\sqrt{b^2-2})^2

    to

    a^2=4(b^2-2)

    Can you explain please?
    (2\sqrt{b^2-2})^2 = 2^2 \cdot (\sqrt{b^2-2})^2 = ?<br />
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  9. #9
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    Quote Originally Posted by Meggomumsie View Post
    Thanks for your answer I understand what you have done, but I don't understand how we get from

    a^2=(2\sqrt{b^2-2})^2

    to

    a^2=4(b^2-2)

    Can you explain please?
    Using a basic index law

    (a\cdot b)^n = a^n\cdot b^n

    We have

    (2\sqrt{b^2 - 2})^2 = 2^2\cdot (\sqrt{b^2 - 2})^2

     = 4(b^2 - 2).
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  10. #10
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    Quote Originally Posted by skeeter View Post
    correction ...

    b^2 - 2 \ge 0

    b \ge \sqrt{2} or b \le -\sqrt{2}

    hence the need for the \pm
    How could I forget that
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