sol-1

    here we have to use  three conditions,

  1)   f(k) > 0                      (K=3 in this question)

 2)   D > = 0

from 1 , f(3) > 0

         a² - 5a + 6 > 0

       (a-3)(a-2) > 0

from this we can say , value of a should be

        a > 3  , a < 2

 from 2 , D> = 0

      4a² > = 4(a²+a-3)

       a-3 < = 0

       a < =  3                  ..................2

these all conditions are satisfied if a < 2

so option  a is correct

ax² + bx +c = 0

in this quadratic expression

 sum of roots = -b/a

        product = c/a

we have eq , 2x² - 3x - 6 = 0

 let x,y are the roots of this eq then

 x+y = -b/a = 3/2

 xy = c/a = -3

x²+y² = (x+y)² - 2xy

x²+y² = (3/2)² - 2(-3) = 9/4+6 = 33/4          ........................1

now , eq whose roots are (x²+2 , y²+2) will be

 [X - (x²+2)][X-(y²+2)] = 0

 X² - X(x²+y²+4) + x²y² + 2(x²+y²) + 4 = 0

x²+y² = 33/4 & xy = -3  so 

X² - X(49/4) + 59/2 = 0

4X² - 49X + 59 = 0

this is the required quadratic equation ...

approve if u like my ans

Dear student,

for real roots b^2 - 4ac > 0

=> 4a^2 - 4a^2 - 4a + 12 > 0

=> a<3

for a = 3 roots of equation are 3,3

so option a is correct.