PDA

View Full Version : Application of quadratic equation

bobbyw
11-04-2008, 02:16 AM
The application goes as follows- "The screen of the TI-84 Plus graphing calculator is nearly rectangular. The length of the rectangle is 2 cm more than the width. If the area of the rectangle is 24 cm[sup:2xfkhg3t]2[/sup:2xfkhg3t].find the length and the width." I know that L x W = Area. I know that given x = the width, than 2x equals the length of the rectangle. I think the equation should go (2x) (x) = 24 cm[sup:2xfkhg3t]2[/sup:2xfkhg3t]. So, given that equation, I use the princple of zero and write, 2x[sup:2xfkhg3t]2[/sup:2xfkhg3t] = 24 and 2x[sup:2xfkhg3t]2[/sup:2xfkhg3t] - 24 = 0. Factoring this out gives me 2(x[sup:2xfkhg3t]2[/sup:2xfkhg3t] - 12). It's at this point that I can't seem to go any further. There's nothing I've used to factor this out anymore and I would like some suggestions, please. Thank you.
I've figured out my problem. It wasn't (2x) (x). It is (2 + x) (x). Thanks anyway :)

mmm4444bot
11-04-2008, 03:40 AM
... 2x[sup:2lvqnwc3]2[/sup:2lvqnwc3] - 24 = 0. Factoring this out gives me 2(x[sup:2lvqnwc3]2[/sup:2lvqnwc3] - 12). It's at this point that I can't seem to go any further. There's nothing I've used to factor this out anymore and I would like some suggestions ...

I'm glad that you figured out the goofup.

As an aside, if you would like to continue solving an equation like your original quadratic by factoring, then you could note that x[sup:2lvqnwc3]2[/sup:2lvqnwc3] - 12 is a difference of squares.

(x + ?12)(x - ?12)

Or, if not by factoring, then continue using algebra to arrive at the following.

x[sup:2lvqnwc3]2[/sup:2lvqnwc3] = 12

|x| = ?12

x = ±?12

Cheers,

~ Mark :)