Question

Find tha nature of root of 2x²-6x+3=0.if real roots exist find them

Answer 1

Answer:

2x2 - 6x + 3 = 0

a = 2, b = - 6, c = 3

b2 - 4ac = (- 6)2 - 4(2)(3)

= 36 - 24

= 12

b2 - 4ac > 0

Hence the equation has two distinct real roots.

We know that, x = [- b ± √ (b2 - 4ac)] / 2a

x = [-(- 6) ± √12] / (2)2

= (6 ± 2√3) / 4

= (3 ± √3) / 2

Roots are x = (3 + √3) / 2 and x = (3 - √3) / 2

Answer 2

Answer:

3±√3/2

Step-by-step explanation: