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6 years ago

510 Points
```										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
```
6 years ago
510 Points
```										ax2 + bx +c = 0
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

```
6 years ago
36 Points
```										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.
```
6 years ago
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