Number of values of ' k text ' so that the equation ( k -3)( k 2-4)( k +1) x 2-( k 3-5 k 2+6 k ) x +( k 2-9)=0 has more than two unequal roots is equal to

Question

Number of values of ' k text ' so that the equation ( k -3)( k 2-4)( k +1) x 2-( k 3-5 k 2+6 k ) x +( k 2-9)=0 has more than two unequal roots is equal to (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

We must have ( k -3)( k 2-4)( k +1)=0, k 3-5 k 2+6 k =0 and k 2-9=0 Hence clearly k =3 only.

Q. Number of values of ' $k^{’}$ so that the equation $(k - 3) (k^{2} - 4) (k + 1) x^{2} - (k^{3} - 5 k^{2} + 6 k) x + (k^{2} - 9) = 0$ has more than two unequal roots is equal to

1

2

3

4

We must have $(k - 3) (k^{2} - 4) (k + 1) = 0, k^{3} - 5 k^{2} + 6 k = 0$ and $k^{2} - 9 = 0$ Hence clearly $k = 3$ only.

Q. Number of values of ' k’ so that the equation (k−3)(k2−4)(k+1)x2−(k3−5k2+6k)x+(k2−9)=0 has more than two unequal roots is equal to

1

2

3

4

We must have (k−3)(k2−4)(k+1)=0,k3−5k2+6k=0 and k2−9=0 Hence clearly k=3 only.

Q. Number of values of ' $k^{’}$ so that the equation $(k - 3) (k^{2} - 4) (k + 1) x^{2} - (k^{3} - 5 k^{2} + 6 k) x + (k^{2} - 9) = 0$ has more than two unequal roots is equal to

We must have $(k - 3) (k^{2} - 4) (k + 1) = 0, k^{3} - 5 k^{2} + 6 k = 0$ and $k^{2} - 9 = 0$ Hence clearly $k = 3$ only.