# Homework Help: Need Help with a Quadratic Equation

1. Sep 16, 2014

### 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~

2. Sep 16, 2014

### Staff: Mentor

Show us what you've done.

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

3. Sep 16, 2014

### 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.

4. Sep 16, 2014

### Staff: Mentor

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

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

5. Sep 16, 2014

### 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.

6. Sep 16, 2014

### Staff: Mentor

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

7. Sep 16, 2014

### Ray Vickson

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

8. Sep 16, 2014

### Staff: Mentor

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

9. Sep 16, 2014

### Ray Vickson

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.

10. Sep 16, 2014

### Staff: Mentor

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

11. Jun 13, 2018

### Merlin3189

Typo aside, it doesn't look much like it to me.
$πr^2 +\left( \frac {25} { πr^2 \times 100 } \right) = π \left( r+6 \right)^2$
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.

12. Jun 13, 2018

### Staff: Mentor

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