Quadratic Equation help

zach610 · #1 $Old$ November 6th, 2009, 02:17 PM

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.

pickslides · #2 $Old$ November 6th, 2009, 02:21 PM

Quote:

Originally Posted by zach610 $View Post$

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

zach610 · #3 $Old$ November 6th, 2009, 02:25 PM

Quote:

Originally Posted by pickslides $View Post$

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.

bigwave · #4 $Old$ November 6th, 2009, 02:25 PM

what calculator are you using

TI89 ???

zach610 · #5 $Old$ November 6th, 2009, 02:27 PM

Quote:

Originally Posted by bigwave $View Post$

what calculator are you using

TI89 ???

TI-84 Plus

I got one of the old ones, without the equation solver

e^(i*pi) · #6 $Old$ November 6th, 2009, 02:30 PM

Quote:

Originally Posted by zach610 $View Post$

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 $10^{24}$

$5.887x^2 - 4.648 \times 10^{9}x + 9.059 \times 10^{17} = 0$

zach610 · #7 $Old$ November 6th, 2009, 02:36 PM

Quote:

Originally Posted by e^(i*pi) $View Post$

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.

e^(i*pi) · #8 $Old$ November 6th, 2009, 02:38 PM

Quote:

Originally Posted by zach610 $View Post$

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 $-\frac{b}{2a}$

zach610 · #9 $Old$ November 6th, 2009, 02:43 PM

Quote:

Originally Posted by e^(i*pi) $View Post$

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.