# Thread: Finding roots a+bi form

1. ## Finding roots a+bi form

Find the roots in simplest a+bi form y=2x^2-4x+5

[Math]\frac{-b+-\sqrt{ b^2-4ac}}{2a}[/tex]

[Math]\frac{4+-\sqrt{16-4(2)(5)}}{2(2)}[/tex]

[Math]\frac{4+-\sqrt{24}}{4}[/tex]

to put in a+bi form I believe that that it has to be a perfect square and that I can not have a radical in the answer? Please help with what I am missing Thanks

2. Originally Posted by IDontunderstand
Find the roots in simplest a+bi form y=2x^2-4x+5

$\displaystyle \frac{-b+-{{Sqrt}b^2-4ac}}{2a}$
As you (kind of) put down the roots of $\displaystyle ax^2+bx+c$ are $\displaystyle x=\frac{b\pm\sqrt{b^2-4ac}}{2a}$

So we can see that the solutions of $\displaystyle \overbrace{2}^{a}x^2-\overbrace{4}^{b}x+\overbrace{5}^{c}$ are $\displaystyle x=\frac{4\pm\sqrt{(-4)^2-4(2)(5)}}{2(2)}=\frac{4\pm\sqrt{16-40}}{4}=\frac{4\pm2\sqrt{6}i}{4}$

3. ## Radical

So am I able to have a radical in simplest a+bi form? I have never seen a radical in simplest form or a fraction which I dont think that should be a problem.

So are we able to have a radical and a fraction in the answer in simple a+bi form?

4. So are we able to have a radical and a fraction in the answer in simple a+bi form?
Technically speaking we should keep going to get
$\displaystyle \frac{4\pm2\sqrt{6}i}{4} = \frac{4}{4}\pm \frac{2\sqrt{6}}{4}i$

then cancel common factors with the denominator, but in reality I would expect the answer given by Mathstud to be worth full marks. Radicals and fractions are perfectly appropriate.