• Apr 1st 2012, 01:13 PM
Past45
 x^2+6x+13=0 -6+-√6^2-4(1)(13)/2(1) -6+-√36-52 -6+-√16/2 -6+-√4/2 -6-4/2=-5 or -6+4/2=-1
Something seems a bit off though.
• Apr 1st 2012, 01:22 PM
emakarov
Quote:

Originally Posted by Past45
x^2+6x+13=0

-6+-√6^2-4(1)(13)/2(1)

You forgot parentheses around -6+-√6^2-4(1)(13).

Quote:

Originally Posted by Past45
-6+-√36-52

Should be (-6+-√36-52)/2

Quote:

Originally Posted by Past45
-6+-√16/2

Again, parentheses around -6+-√16 are missing. Most importantly, 36 - 52 ≠ 16.
• Apr 1st 2012, 01:26 PM
Plato
Quote:

Originally Posted by Past45
 x^2+6x+13=0 -6+-√6^2-4(1)(13)/2(1) -6+-√36-52 -6+-√16/2 -6+-√4/2 -6-4/2=-5 or -6+4/2=-1
Something seems a bit off though.

It is way off. The roots are complex numbers: $\pm\sqrt{36-52}=\pm 4\bf{i}$
• Apr 1st 2012, 01:29 PM
Past45
Hu? Sorry a little confused. I did put the parentheses.. Only time in the book the parentheses are put in are when I plug in my numbers.

The answer for this is suppose to -3-2i or -3+2i
• Apr 1st 2012, 01:29 PM
emakarov
Quote:

Originally Posted by Plato
The roots are complex numbers: $\pm\sqrt{36-52}=\pm 4\bf{i}$

You must mean the square roots of the discriminant, not the roots of the equation.

• Apr 1st 2012, 01:38 PM
Past45
-6 ± √4i
----------- =
2

2(-3 √4i)
----------- =
2

the 2 cancels out so it will -3+-√2i

so will it look like this?
• Apr 1st 2012, 01:39 PM
emakarov
Quote:

Originally Posted by Past45
I did put the parentheses.. Only time in the book the parentheses are put in are when I plug in my numbers.

The roots of the equation are $\frac{-6\pm\sqrt{6^2-4\cdot1\cdot13}}{2\cdot 1}$, which you wrote as -6+-√6^2-4(1)(13)/2(1). As you know, in the standard order of operations division comes before subtraction, so your formula is $-6\pm\sqrt{6^2-\frac{4\cdot1\cdot13}{2\cdot1}$. (In the text formula, it is also not clear where the square root ends.) And what about -6+-√16/2, which means $-6\pm\sqrt{\frac{16}{2}}$ or $-6\pm\frac{\sqrt{16}}{2}$, but not $\frac{-6\pm\sqrt{16}}{2}$?

In any case, the main mistake is that the discriminant is negative.
• Apr 1st 2012, 01:41 PM
Past45
Nevermind I kind of see it, -6/2 =-3. Duh..
• Apr 1st 2012, 01:42 PM
Past45
thanks.

Quote:

Originally Posted by emakarov
The roots of the equation are $\frac{-6\pm\sqrt{6^2-4\cdot1\cdot13}}{2\cdot 1}$, which you wrote as -6+-√6^2-4(1)(13)/2(1). As you know, in the standard order of operations division comes before subtraction, so your formula is $-6\pm\sqrt{6^2-\frac{4\cdot1\cdot13}{2\cdot1}$. (In the text formula, it is also not clear where the square root ends.) And what about -6+-√16/2, which means $-6\pm\sqrt{\frac{16}{2}}$ or $-6\pm\frac{\sqrt{16}}{2}$, but not $\frac{-6\pm\sqrt{16}}{2}$?

In any case, the main mistake is that the discriminant is negative.

• Apr 1st 2012, 01:43 PM
emakarov
Quote:

Originally Posted by Past45
-6 ± √4i

√4i should be replaced by i√(16)
• Apr 1st 2012, 01:47 PM
Plato
Quote:

Originally Posted by emakarov
You must mean the square roots of the discriminant, not the roots of the equation.

I did not say that those were the roots.
I meant what I said.
• Apr 1st 2012, 01:49 PM
Past45
I am just going by how the book it showing me how to do it. But the square root of 16, is 4 though.

Quote:

Originally Posted by emakarov
√4i should be replaced by i√(16)

• Apr 1st 2012, 02:04 PM
emakarov
Quote:

Originally Posted by Past45
I am just going by how the book it showing me how to do it. But the square root of 16, is 4 though.

Yes, and the equation roots are $\frac{-6\pm4i}{2}=-3\pm2i$.

You should pay much more attention to arithmetic. Even though you are claiming that you are doing everything by the book, your posts in this thread contain numerous arithmetic errors.
• Apr 1st 2012, 02:05 PM
Past45
I just got it done, it is a lot simpler. The little errors are throwing me off. Thanks ne who.

And yes, I go over the book and notes before anything. Thanks again.

Quote:

Originally Posted by emakarov
Yes, and the equation roots are $\frac{-6\pm4i}{2}=-3\pm2i$.

You should pay much more attention to arithmetic. Even though you are claiming that you are doing everything by the book, your posts in this thread contain numerous arithmetic errors.

• Apr 1st 2012, 05:38 PM
Wilmer