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In the following equation how would I rewrite the equation to solve for $m$?

$$z=\frac{-4m-8+\sqrt{(4m+8)^2+4(4(mx+y-4m-4))}}{8}$$

when $x=66$ and $y=22$ and $z=10$

DeveloperChris
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  • If you look closely you can see that the right hand operator is a solution for a quadratic equation where b=4m+8, a=4 and c=mx+y-4m-4. Then use Viete's formulas to create your equation system. z is one of the solutions. To find the other solution of the quadratic equation just change the sign before the sqrt to - – wxyz Dec 13 '13 at 13:26
  • I am fully aware this is a reworked quadratic formula. – DeveloperChris Dec 13 '13 at 22:00
  • I had a quick read about vieta's formula's in wikipedia, although I can see that they apply to quadratic equations I am afraid it is well above me :(. If I was at that level I don't think I'd be asking questions here. rather I'd be answering them. – DeveloperChris Dec 13 '13 at 22:34

2 Answers2

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the given expression is of the form $$z=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ which can be reordered to form a quadratic in$z$. $$az^2+bz+c=0$$ $a=4$,$b=4m+8$,$c=-(mx+y-4m-4)$

Suraj M S
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  • Your guidance has confused me somewhat but I'll sit and think through what you have written. As I know neither b or c without first knowing m I am not sure I understand how your answer helps solve for m. – DeveloperChris Dec 13 '13 at 22:40
  • how do you solve a quadratic equation to find its roots. this is just its inverse. ie to construct a quadratic equation so that finding a root to this equation gives you the expression in your question. – Suraj M S Dec 14 '13 at 03:24
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$(1)$. Multiply both sides with $8$ ; $(2)$. Add $4m+8$ to both sides. $(3)$. Square both sides.

$(4)$. Subtract $(4m+8)^2$ from both sides. $(5)$. Divide both sides through $16$. You'll get:

$$4z^2+z\,(4m+8)=m(x-4)+(y-4)$$

Now, expand the parentheses on both sides, move all terms containing m to one side, and all other terms to the other side, then divide through the factor of m, and you're done.

Lucian
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  • Thank you. I haven't had time to work this through yet but will later tonight. It would seem you have answered my question, now I have to see if I asked the right question ;) – DeveloperChris Dec 13 '13 at 22:45