Question

plz help.........plz ..............​

Answer 1

Answer:

Here, a= √3 b=-1/√3&c=-4/√3

So, = -b+_√b^2-4ac/2a

= {-(-1/√3)+_√(-1/√3)^2 -4*√3*(-4/3)}/2√3

={1/√3+_√1/3-4*(-4)/}2√3

={1/√3+_√15}/2√3

=( 1/√3+_√5)/2√3

= (1+_√5/√3)/2√3

= 6/1+_√5

So, two roots of the given quadratic equation are:

6/1+√5 & 6/2-√5

Hope it helps you...

Answer 2

Answer:

Thank you sooo much whatever your name is. Thanks a lot.