Converting a formula into quadratic equation

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Simonsky

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I'm using the formula for surface area of a cylinder: 2(Pi)r^2 +2(Pi)rh.

Question asks: tin of height 6cm has surface area 54(pi)cm^2 find radius.

So I first created this quadratic: 2(Pi)r^2 + 12(Pi)r = 54(Pi)cm^2

Then 2(Pi)r^2 + 12(Pi)r - 54(Pi)cm2 = 0

Not sure how to go from here. Does one use substitution or use the Pi +coefficient as 'a' and 'b' in the quadratic formula ax^2 + bX +c =0

No need to do the whole thing for me-I jsut need a nudge.

Thanks
 
Simonsky said:
I'm using the formula for surface area of a cylinder: 2(Pi)r^2 +2(Pi)rh.

Question asks: tin of height 6cm has surface area 54(pi)cm^2 find radius.

So I first created this quadratic: 2(Pi)r^2 + 12(Pi)r = 54(Pi)

Then 2(Pi)r^2 + 12(Pi)r - 54(Pi) = 0

Not sure how to go from here. Does one use substitution or use the Pi +coefficient as 'a' and 'b' in the quadratic formula ax^2 + bX +c =0

No need to do the whole thing for me-I jsut need a nudge.

Thanks
Click to expand...
2(Pi)r^2 + 12(Pi)r - 54(Pi) = 0 ................................. should not have cm2 here

\(\displaystyle 2 * \pi\) * r^2 + 12 * \(\displaystyle \pi\) * r - 54 * \(\displaystyle \pi\) = 0 ......................... divide through by 2 * \(\displaystyle \pi\)

r^2 + 6 * r - 27 = 0

ax^2 + bx + c = 0 ......................................be consistent - use either 'X' or 'x'

a = 1, b = 6 & c = -27
 
Thanks.

So I use this: r^2 +6r - 27 = 0

The factorise: r^2 +9r - 3r -27 = 0

Then: r(r+9) -3(r+9) = 0 giving: (r-3)(r+9) so r=3 and r= -9. In the question it must be a positive number so r =3cm.

What puzzles me is how a quadratic can have two roots yet only a positive one is admissible in this case?
 
Simonsky said:
Thanks.

So I use this: r^2 +6r - 27 = 0

The factorise: r^2 +9r - 3r -27 = 0

Then: r(r+9) -3(r+9) = 0 giving: (r-3)(r+9) so r=3 and r= -9. In the question it must be a positive number so r =3cm.

What puzzles me is how a quadratic can have two roots yet only a positive one is admissible in this case?
Click to expand...
"Mathematically" two possible answers - "physically" one possible answer.
 
But maybe there is a parallel world of negative heights? (Underground or underwater structures!) Maths seems to allow all sorts of imaginary and transcendental thingso_O Not that I have the cognitive skills to comprehend that, of course!
 
Given such a parallel world, did this problem refer to it? Sometimes you just use common sense.
 
Just me musing, nothing more than that-is one not allowed to roam around imaginatively even if the question is prosaic and not in that realm?
 
You are welcome to "roam" around as you please! But you wrote it as a response to this specific question. Did it help to answer this question?
 
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