Question

values of x that are roots of the polynomial. x^2-4x+6.

Answer 1

Does the quadratic factor?

Answer 2

yes I think so

Answer 3

it'll have complex roots

Answer 4

like square roots? I figured. Which is why I need help

Answer 5

So, you'll want to use the quadratic formula.

Answer 6

I know but I still cant figure it out

Answer 7

sorry, browser crashed...

Answer 8

its ok

Answer 9

so , a = 1, b = -4, c = 6,
\[x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}\]
\[x = \frac{-(-4) \pm \sqrt{(-4)^2 -4(1)(6)}}{2(1)}\]

Answer 10

a little simplifying,
\[x = \frac{4 \pm \sqrt{16 -24}}{2}\]
\[x = \frac{4 \pm \sqrt{-8}}{2}\]

Answer 11

then
\[\sqrt{-8} = i\sqrt{8} = i\sqrt{4 \cdot 2} = 2i\sqrt{2}\]

Answer 12

So simplifying some more,
\[x = \frac{4 \pm 2i\sqrt{2}}{2}\]
Now we can factor 2 out of the numerator, and cancel it with the 2 in the denominator,
\[x = \frac{2(2 \pm i\sqrt{2})}{2}\]
\[x = \frac{2 \pm i\sqrt{2}}{1}\]
\[x = 2 \pm i\sqrt{2}\]