The range of values of a for which roots of the quadratic equation x^2+x-3a+2=0 are of opposite sign is given bya) a<1b) a>2c) 1

Your given equation is

x² + x – 3a + 2 = 0

Since the roots are opposite in sign, so α x β is –ve                  (i)

Hence,

α x β = -3a + 2/1                 (ii)

Therefore, product of roots of the eq.

ax² + bx + c = c/a

From equation (i) and (ii), we have

-3a + 2 < 0

-3a < -2

a > 2/3                           (i)

Also, discriminant D should be greater than 0 for roots to be real

Hence, 1 – 4 (-3a + 2) > 0

1 > 4(-3a + 2)

-12a + 8 < 1

-12a < -7

a > 7/12                          (ii)

Taking the intersection (i) and (ii) we have,

a > 2/3

Hence, the most appropriate option is (b) a > 2 , since a > 2/3

Then, a > 2.