Question

Help question in comments!

Answer 1

First, can you put it in the form to use the quadratic formula?

Answer 2

\[\frac{4\pm \sqrt{(-4)^2-4*1*1} }{ 2*1}\]

Answer 3

wait, one step earlier please

Answer 4

In order to use the quadratic formula, your eq must look like this: \(ax^2+bx+c=0\)

Answer 5

x^2-4x+1=0?

Answer 6

yep ok, so now, what are your a,b,c=?

Answer 7

a=1 b=-4 c=1

Answer 8

Good, and can you state the quadratic formula?

Answer 9

\[\frac{ -bpm \sqrt{b-4ac} }{ 2a }\]

Answer 10

\[\pm\] not pm sorry

Answer 11

ok I think you meant b^2 under correct?

Answer 12

yes sorry

Answer 13

I understood that haha no probs

Answer 14

so you subbed in correctly, but just to be picky, I would include parentheses to avoid confusion: \[x=\frac{-(-4) \pm \sqrt{(-4)^2-(4*1*1)} }{ 2*1}\]

Answer 15

ok

Answer 16

So now, let's simplify as much as you can, leave the square root as is for now

Answer 17

\[\frac{ 2\sqrt{(-4)^2-(4*1*1)} }{ 3 }\]

Answer 18

not quite ok how, did you get that?

Answer 19

oh wow never mind ignore that

Answer 20

what do you mean simplify but leave the square root?

Answer 21

@FibonacciChick?

Answer 22

@ParthKohli ?

Answer 23

@Erob ... see... use quardratic formula... here according to your equation... a= 1... b= -4... c= 1... by putting all the vlaues in the equation... we will get... (2±√3)... so option A...

Answer 24

Thank you!

Answer 25

pleasure