2
$\begingroup$

I need to re-arrange the following equation to make $d$ the subject [i.e., solve for $d$]:

$$a = \sqrt{\frac{d}{be}}.$$

The square root part is really confusing me, any help would be much appreciated.

  • 0
    @Dennis: What you mean, when you say : Make $d$ the subject2010-11-18
  • 0
    So as the equation is re-arranged to be d =2010-11-18

1 Answers 1

2

Ok. Squaring both sides you obtain $$a^{2} = \frac{d}{be} \Longrightarrow d = (be) \times a^{2}$$

  • 0
    Thanks, is it possible to re-arrange it without the brackets though?2010-11-18
  • 0
    @Dennis: Yes, i used brackets only for the sake of convenience2010-11-18
  • 0
    Thanks, so what has happened to the 'a' part?2010-11-18
  • 0
    @Dennis: According to your notation $a=M^{2}b^{2}e^{2}$2010-11-18
  • 0
    Sorry, I'm not sure if I'm managed to explain it correctly, please can you see the below image of the equation?2010-11-18
  • 0
    Saved as a JPG image: http://img42.imageshack.us/img42/7428/equation.jpg I need to make the the 'd' the subject.2010-11-18
  • 0
    @dennis: I have edited the answer now.2010-11-18
  • 0
    If you want extra points for precision, note down “if $\Re a>0$” after your answer. For additional credit, extend that condition to the full set of $a$ for which $a=\sqrt{c}$ has an answer, it's very similar to the condition above …2010-11-19