Question

2. x²-6x+1=0
would the answer to this be
x=-3+√10 AND x=-3-√10

Answer 1

that is not correct

Answer 2

can you help, I have no idea how to do it

Answer 3

x^2 - 6x + 1 = 0 is in the form ax^2 + bx + c = 0

where

a = 1
b = -6
c = 1

Answer 4

see how I'm identifying a,b,c?

Answer 5

yeah, I already did that

Answer 6

now you plug those values into the quadratic formula

\[\Large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 7

So we get this
\[\Large x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[\Large x = \frac{-(-6) \pm \sqrt{(-6)^2-4(1)(1)}}{2(1)}\]
What's next?

Answer 8

got that, I got to -(-6)+-√36-(4)
----------------------------
2

Answer 9

then you'll have \[\Large x = \frac{6 \pm \sqrt{32}}{2}\]

Answer 10

now you have to simplify \[\Large \sqrt{32}\]

Answer 11

5.6?

Answer 12

You simplify like this
\[\Large \sqrt{32} = \sqrt{16*2}\]
\[\Large \sqrt{32} = \sqrt{16}*\sqrt{2}\]
\[\Large \sqrt{32} = 4\sqrt{2}\]

Answer 13

so... x=4\[\sqrt{2}\]

Answer 14

not quite

Answer 15

\[\Large x = \frac{6 \pm \sqrt{32}}{2}\]
will turn into
\[\Large x = \frac{6 \pm 4\sqrt{2}}{2}\]

Answer 16

x=10√2 ?

Answer 17

you cannot add the 6 and 4 (since the 4 has a root 2 attached to it)

Answer 18

what you can do though is factor out 2 from the numerator, then divide like this

\[\Large x = \frac{6 \pm 4\sqrt{2}}{2}\]
\[\Large x = \frac{2(3 \pm 2\sqrt{2})}{2}\]
\[\Large x = 3 \pm 2\sqrt{2}\]

Answer 19

so, x=3+√2 and x=3-√2?

Answer 20

I would replace the 'and' with 'or' since x cannot be two values at once, but yes, the solutions are

x=3+√2 or x=3-√2

Answer 21

thanks so much!

Answer 22

you're welcome