• Nov 6th 2009, 01:17 PM
zach610
For some reason my calculator wont seem to work this out...help please

$0 = 5.887E24x^2 - 4.648E33x + 9.059E41$

The "E"s are meant to represent scientific notation. You could replace them with a "5.887x10^5" type of thing, but I thought that would have made it more confusing.

Yes, I have tried the quadratic equation (the negative "b" plus or minus the square root of "b" squared minus BLAH BLAH BLAH). I know the equation by heart, and tried it in my calculator. Could someone please help with this.
• Nov 6th 2009, 01:21 PM
pickslides
Quote:

Originally Posted by zach610

Yes, I have tried the quadratic equation (the negative "b" plus or minus the square root of "b" squared minus BLAH BLAH BLAH). I know the equation by heart, and tried it in my calculator. Could someone please help with this.

Then you should have the answer
• Nov 6th 2009, 01:25 PM
zach610
Quote:

Originally Posted by pickslides
Then you should have the answer

I tried that, but it gives me the same answer with the plus, and the minus. When, instead it is supposed to give you one answer negative, and the other is positive.
• Nov 6th 2009, 01:25 PM
bigwave
what calculator are you using
what calculator are you using

TI89 ???
• Nov 6th 2009, 01:27 PM
zach610
Quote:

Originally Posted by bigwave
what calculator are you using

TI89 ???

TI-84 Plus

I got one of the old ones, without the equation solver
• Nov 6th 2009, 01:30 PM
e^(i*pi)
Quote:

Originally Posted by zach610
I tried that, but it gives me the same answer with the plus, and the minus. When, instead it is supposed to give you one answer negative, and the other is positive.

Not always. If $b^2-4ac=0$ then the roots will be real and equal

$0 = 5.887 \times 10^{24}x^2 - 4.648 \times 10^{33}x + 9.059 \times 10^{41} = 0$

Divide by $10^{24}$

$5.887x^2 - 4.648 \times 10^{9}x + 9.059 \times 10^{17} = 0$
• Nov 6th 2009, 01:36 PM
zach610
Quote:

Originally Posted by e^(i*pi)
Not always. If $b^2-4ac=0$ then the roots will be real and equal

Hm...okay. Well, could some one try this and tell me what you get. I got "x" to be 4.648E33 for both the negative and the positive answer. This is also one of my numbers up there...something seems wrong. This is why I came here, because I am almost sure I am doing something wrong.
• Nov 6th 2009, 01:38 PM
e^(i*pi)
Quote:

Originally Posted by zach610
Hm...okay. Well, could some one try this and tell me what you get. I got "x" to be 4.648E33 for both the negative and the positive answer. This is also one of my numbers up there...something seems wrong. This is why I came here, because I am almost sure I am doing something wrong.

I am guessing because $b >> b^2-4ac$ then you can assume $b^2-4ac=0$ which would give a root equal to $-\frac{b}{2a}$
• Nov 6th 2009, 01:43 PM
zach610
Quote:

Originally Posted by e^(i*pi)
I am guessing because $b >> b^2-4ac$ then you can assume $b^2-4ac=0$ which would give a root equal to $-\frac{b}{2a}$

Well, when I try that, I get a completely different number. This is exactly why I posted this. This problem is much harder than I thought. Not sure where to go from here.