Question

can anyone please solve it . ​

Answer 1

Answer:

i) No Real Roots

ii) Equal Roots; 2/√3, 2/√3

iii) Distinct Roots; [3±√3]/2

Answer 2

1)2x²-3x+5=0

=> Given equation ax²+bx+c=5,in which a=2,b= -3,c= 5

=> Therefore D= b²-4ac

=> (-3)-4×2×5= 9-40= -31<0

Hence,There is no real roots for this quadratic equation..

2) 3x²-4√3x+4=0

=> Given equation ax²+bx+c=0,in which a=3,b= -4√3,c=4

Therefore D=b²-4ac

=> (-4√3)²-4×3×4

=> 48-48=0

So,the roots of quadratic equation are real and equal.

Hence,x=4√3±√0/6. [As x= -b±√b²-4ac/2a]

=> 4√3/6= 2√3/6

Hence,the roots of the quadratic equation are 2√3/3 and 2√3/3..

(3) 2x²-6x+3=0

=> Given equation ax²+bc+c=0,in which a=2,b= -6,c=3

Therefore D= b²-4ac

=> (-6)-4×2×3= 36-24=12>0

Hence,x=6±√12/4= 6±2√3/4= 3±√3/2

either,x=3±√3/2 or x=3-√3/2

Hence,the roots of the quadratic equation are 3+√3/2 and 3-√3/2..

It was so Easy Question.....