# Solutions for imaginary numbers in a+bi

• Jan 30th 2010, 07:25 PM
DustinP
Solutions for imaginary numbers in a+bi
A sample problem is:

Find all solutions of the equation and express them in the form a + bi.

x^2-7x+13=0

I'm supposed to get 2 answers. A positive conjugate and a negative conjugate.

However when I solve it, I get an odd fraction and the website keeps telling me that this is a wrong answer. And the formatting is right. So I know I'm doing something wrong.

Could someone tell me how these sort of problems are solved? I know most of the steps, but I'm missing something.

Thanks.
• Jan 30th 2010, 07:31 PM
Prove It
Quote:

Originally Posted by DustinP
A sample problem is:

Find all solutions of the equation and express them in the form a + bi.

x^2-7x+13=0

I'm supposed to get 2 answers. A positive conjugate and a negative conjugate.

However when I solve it, I get an odd fraction and the website keeps telling me that this is a wrong answer. And the formatting is right. So I know I'm doing something wrong.

Could someone tell me how these sort of problems are solved? I know most of the steps, but I'm missing something.

Thanks.

You can complete the square or use the Quadratic Formula:

$\displaystyle x^2 - 7x + 13 = 0$

$\displaystyle x^2 - 7x + \left(-\frac{7}{2}\right)^2 - \left(-\frac{7}{2}\right)^2 + 13 = 0$

$\displaystyle \left(x - \frac{7}{2}\right)^2 - \frac{49}{4} + \frac{52}{4} = 0$

$\displaystyle \left(x - \frac{7}{2}\right)^2 + \frac{3}{4} = 0$

$\displaystyle \left(x - \frac{7}{2}\right)^2 = -\frac{3}{4}$

$\displaystyle x - \frac{7}{2} = \pm \sqrt{-\frac{3}{4}}$

$\displaystyle x - \frac{7}{2} = \frac{\pm \sqrt{-3}}{\sqrt{4}}$

$\displaystyle x - \frac{7}{2} = \frac{\pm \sqrt{3}i}{2}$

$\displaystyle x = \frac{7}{2} \pm \frac{\sqrt{3}}{2}i$

So the two solutions are:

$\displaystyle x = \frac{7}{2} + \frac{\sqrt{3}}{2}i$ and $\displaystyle x = \frac{7}{2} - \frac{\sqrt{3}}{2}i$.
• Jan 31st 2010, 02:36 AM
HallsofIvy
Quote:

Originally Posted by DustinP
A sample problem is:

Find all solutions of the equation and express them in the form a + bi.

x^2-7x+13=0

I'm supposed to get 2 answers. A positive conjugate and a negative conjugate.

However when I solve it, I get an odd fraction and the website keeps telling me that this is a wrong answer. And the formatting is right. So I know I'm doing something wrong.

It would have been better if you had shown us what wrong answer you got!

Quote:

Could someone tell me how these sort of problems are solved? I know most of the steps, but I'm missing something.

Thanks.
• Jan 31st 2010, 03:16 PM
DustinP
Ahh. Ok Thanks. I put that answer in and it is right.

What I was doing (or not doing) was keeping the square root of 3. So I just had 3 over 2. But I see what you did now, I think.

• Jan 31st 2010, 03:19 PM
DustinP
$\displaystyle x = \frac{7}{2} - \frac{3}{2}i$
$\displaystyle x = \frac{7}{2} + \frac{3}{2}i$
$\displaystyle x=\frac{7\pm\sqrt{7^2-4(13)}}{2}$