How Do You Solve a Circle's Radius Problem Using the Quadratic Formula?

[Saracen Rue]

Hi, first off I want to say that I'm new here, so sorry if I do anything wrong.

Okay, now to the problem at hand. I know that this is probably really easy and I'm just having one of my moments again, but I can't for the life of me figure out how to do this question:

When the radius of a circle is increased by 6 cm, its area increases by 25%. Use the quadratic formula to find out the exact radius of the original circle.

I've spent a good 2-3 hours on it and I just can't seem to get it into the ax²+bx+c form. Any help and insight into this question you guys could provide me would be much appreciated. Thank you all for your time.

~Saracen Rue~

 

[Doc Al]

Show us what you've done.

Hint: What's the formula for the area of a circle?

 

[Saracen Rue]

Well, the first thing I did was assign area formulas to both the original circle and the new circle, with the original one being A = πr² and the new one being A = π(r+6)². Now I needed to make the equations equal each other, which I did by adding 25% of the area of the first circle onto itself, giving me something that looked like this: πr² + (25/πr²*100) = πr(r+6)². Next, I moved the πr(r+6)² over to the left hand side to make the equation equal to 0, allowing me to treat it as a quadratic equation. However, here's where I get stuck. I can't work out how to make it into the form ax²+bx+c.

 

[Doc Al]

Well, the first thing I did was assign area formulas to both the original circle and the new circle, with the original one being A = πr² and the new one being A = π(r+6)². Now I needed to make the equations equal each other, which I did by adding 25% of the area of the first circle onto itself, giving me something that looked like this: πr² + (25/πr²*100) = πr(r+6)².

Good, but you have an extra r in that right hand term. Get rid of it.

Next, I moved the πr(r+6)² over to the left hand side to make the equation equal to 0, allowing me to treat it as a quadratic equation. However, here's where I get stuck. I can't work out how to make it into the form ax²+bx+c.

Cancel what you can cancel. Then be sure to expand that (r + 6)² term.

 

[Saracen Rue]

Yeah, that r was a typo.

Hm, I think my problem is that I keep getting confused with my negatives -_-
and I think I might just be tired... it is 12 in the morning where I am. I think I'll go get some sleep and try again once I'm rested. Good night.

 

[Doc Al]

Sleep is good. But you're on the right track.

 

[Ray Vickson]

Well, the first thing I did was assign area formulas to both the original circle and the new circle, with the original one being A = πr² and the new one being A = π(r+6)². Now I needed to make the equations equal each other, which I did by adding 25% of the area of the first circle onto itself, giving me something that looked like this: πr² + (25/πr²*100) = πr(r+6)². Next, I moved the πr(r+6)² over to the left hand side to make the equation equal to 0, allowing me to treat it as a quadratic equation. However, here's where I get stuck. I can't work out how to make it into the form ax²+bx+c.

Why don't you just write $1.25 \pi r^2 = \pi (r+6)^2$? That is exactly what the question says!

 

[Doc Al]

Why don't you just write $1.25 \pi r^2 = \pi (r+6)^2$? That is exactly what the question says!

That's just what the OP did! (Although it might not look like it.)

 

[Ray Vickson]

That's just what the OP did! (Although it might not look like it.)

Yes, I know that. But he should develop a habit of writing things more compactly. It may even be that different ways of writing correspond to different ways of thinking, so that is why I wrote what I did.

 

[Doc Al]

Yes, I know that. But he should develop a habit of writing things more compactly. It may even be that different ways of writing correspond to different ways of thinking, so that is why I wrote what I did.

I certainly agree that it should be written as you wrote it.

 

[Merlin3189]

That's just what the OP did! (Although it might not look like it.)

Typo aside, it doesn't look much like it to me.

πr2 + (25/πr2*100) = πr(r+6)2.

$πr^2 +\left( \frac {25} { πr^2 \times 100 } \right) = π \left( r+6 \right)^2$
I suppose it could be
$πr^2 +\left( \frac {25} { πr^2} \times 100 \right) = π \left( r+6 \right)^2$
but that's no better

He didn't show his working, but perhaps he didn't think what he wrote was the same as Ray wrote. Then he'd get something more complex and have trouble solving it. So it may not help to tell him his expression is right.

I know this is long after the thread is closed and of no use to OP. He just posted a strange math question and I was looking back through his posts to see where he was coming from.
When I read this thread, I thought this was a point worth making.

 

[Chestermiller]

Well, the first thing I did was assign area formulas to both the original circle and the new circle, with the original one being A = πr² and the new one being A = π(r+6)². Now I needed to make the equations equal each other, which I did by adding 25% of the area of the first circle onto itself, giving me something that looked like this: πr² + (25/πr²*100) = πr(r+6)².

The equation should correctly read: $$\pi r^2+25*\pi r^2/100=\pi (r+6)^2$$
This leads directly to Ray's equation:
$$1.25 \pi r^2=\pi(r+6)^2$$