Quadratic Equation via Formula

| 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.

Re: Quadratic Equation via Formula

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.

Re: Quadratic Equation via Formula

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: $\displaystyle \pm\sqrt{36-52}=\pm 4\bf{i}$

Re: Quadratic Equation via Formula

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

Re: Quadratic Equation via Formula

Quote:

Originally Posted by

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

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

Edit: added quote.

Re: Quadratic Equation via Formula

-6 ± √4i

----------- =

2

2(-3 √4i)

----------- =

2

the 2 cancels out so it will -3+-√2i

so will it look like this?

Re: Quadratic Equation via Formula

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 $\displaystyle \frac{-6\pm\sqrt{6^2-4\cdot1\cdot13}}{2\cdot 1}$, 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 $\displaystyle -6\pm\sqrt{6^2-\frac{4\cdot1\cdot13}{2\cdot1}$. (In the text formula, it is also not clear where the square root ends.) And what about -6+-√16/2, which means $\displaystyle -6\pm\sqrt{\frac{16}{2}}$ or $\displaystyle -6\pm\frac{\sqrt{16}}{2}$, but not $\displaystyle \frac{-6\pm\sqrt{16}}{2}$?

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

Re: Quadratic Equation via Formula

Nevermind I kind of see it, -6/2 =-3. Duh..

Re: Quadratic Equation via Formula

thanks.

Quote:

Originally Posted by

**emakarov** The roots of the equation are $\displaystyle \frac{-6\pm\sqrt{6^2-4\cdot1\cdot13}}{2\cdot 1}$, 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 $\displaystyle -6\pm\sqrt{6^2-\frac{4\cdot1\cdot13}{2\cdot1}$. (In the text formula, it is also not clear where the square root ends.) And what about -6+-√16/2, which means $\displaystyle -6\pm\sqrt{\frac{16}{2}}$ or $\displaystyle -6\pm\frac{\sqrt{16}}{2}$, but not $\displaystyle \frac{-6\pm\sqrt{16}}{2}$?

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

Re: Quadratic Equation via Formula

Quote:

Originally Posted by

**Past45** -6 ± √4i

√4i should be replaced by i√(16)

Re: Quadratic Equation via Formula

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.

Re: Quadratic Equation via Formula

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)

Re: Quadratic Equation via Formula

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 $\displaystyle \frac{-6\pm4i}{2}=-3\pm2i$.

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.

Re: Quadratic Equation via Formula

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 $\displaystyle \frac{-6\pm4i}{2}=-3\pm2i$.

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.

Re: Quadratic Equation via Formula

Quote:

Originally Posted by

**Past45** x^2+6x+13=0

-6+-√6^2-4(1)(13)/2(1)

Properly bracketed (and you should show as equation):

x = {-6 +- √[6^2 -4(1)(13)]} / [2(1)]