Results 1 to 10 of 10

Thread: Another change the subject question

  1. #1
    Junior Member
    Joined
    Aug 2009
    From
    Derbyshire, central UK
    Posts
    28

    Another change the subject question

    Not sure where to start with this one

    $\displaystyle a=2\sqrt{b^2-2}$ (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; Sep 27th 2009 at 08:15 AM. Reason: Forgot to identify the new subject for the formula
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by Meggomumsie View Post
    Not sure where to start with this one

    $\displaystyle a=2\sqrt{b^2-2}$ (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.
    $\displaystyle a = 2\sqrt{b^2 - 2}$

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

    $\displaystyle a^2 = 4(b^2 - 2)$

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

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

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

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

    $\displaystyle b = \frac{\sqrt{a^2 + 8}}{2}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by Prove It View Post

    $\displaystyle b = \textcolor{red}{\pm}\frac{\sqrt{a^2 + 8}}{2}$.
    ...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    Quote Originally Posted by skeeter View Post
    ...
    Of course...

    I didn't want to give EVERYTHING away
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by skeeter View Post
    ...
    $\displaystyle b > \sqrt{2}$

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

    unless the OP is including the complex numbers
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Aug 2009
    From
    Derbyshire, central UK
    Posts
    28
    Thanks for your answer I understand what you have done, but I don't understand how we get from

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

    to

    $\displaystyle a^2=4(b^2-2)$

    Can you explain please?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    Quote Originally Posted by e^(i*pi) View Post
    $\displaystyle b > \sqrt{2}$
    correction ...

    $\displaystyle b^2 - 2 \ge 0$

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

    hence the need for the $\displaystyle \pm$
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    16,216
    Thanks
    3702
    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

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

    to

    $\displaystyle a^2=4(b^2-2)$

    Can you explain please?
    $\displaystyle (2\sqrt{b^2-2})^2 = 2^2 \cdot (\sqrt{b^2-2})^2 = ?
    $
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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

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

    to

    $\displaystyle a^2=4(b^2-2)$

    Can you explain please?
    Using a basic index law

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

    We have

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

    $\displaystyle = 4(b^2 - 2)$.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by skeeter View Post
    correction ...

    $\displaystyle b^2 - 2 \ge 0$

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

    hence the need for the $\displaystyle \pm$
    How could I forget that
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Dec 28th 2011, 11:47 AM
  2. Replies: 2
    Last Post: Sep 7th 2010, 12:04 PM
  3. inverse Kinematics/change of subject formular
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: Dec 30th 2009, 11:39 PM
  4. Change subject
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Nov 17th 2009, 07:49 PM
  5. Changing the subject question help!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 15th 2007, 10:28 AM

Search Tags


/mathhelpforum @mathhelpforum