Question

Root 2x2-3x+1 divided by root 2=0
Find the nature of roots on the following quadratic equation

Answer 1

=> √2x²-3x+1 /√2 = 0 => 2x²-3x+1 /2=0 =>2x²+(-3x)+1 =0 now comparing with ax²+bx + c =0 ; a=2 , b = 3 , c =1 .

=> √2x²-3x+1 /√2 = 0 => 2x²-3x+1 /2=0 =>2x²+(-3x)+1 =0 now comparing with ax²+bx + c =0 ; a=2 , b = 3 , c =1 . now ,D = b²-4ac= 9-8=1 ..°•° D>0 ..•°• real and distinct roots ..verification ::

real and distinct roots ..verification :: if u mean this :

real and distinct roots ..verification :: if u mean this : => √2x²-3x+1 /√2 = 0 => 2x²-3x+1 /2=0 =>2x²-3x+1 =0 => 2x²-2x-x +1 =0 =>2x(x-1)-1(x-1) =0

real and distinct roots ..verification :: if u mean this : => √2x²-3x+1 /√2 = 0 => 2x²-3x+1 /2=0 =>2x²-3x+1 =0 => 2x²-2x-x +1 =0 =>2x(x-1)-1(x-1) =0 now case 1: => 2x-1=0 =>x= 1/2

real and distinct roots ..verification :: if u mean this : => √2x²-3x+1 /√2 = 0 => 2x²-3x+1 /2=0 =>2x²-3x+1 =0 => 2x²-2x-x +1 =0 =>2x(x-1)-1(x-1) =0 now case 1: => 2x-1=0 =>x= 1/2 case 2 : => x=1