Question

Find the values of the parameter k for which the equation x4 - (k - 3) x2 + k = 0 has
(i) Four real roots
(ii) Exactly two real roots
(iii) No real root​

Answer 1

Answer:

plz mark me as brainlist

Answer 2

Given:  x4 - (k - 3) x2 + k = 0

To Find: value of k

Solution:

The equation $x^{4}$ - (k-3)$x^{2}$ + k = 0 can be transformed to $t^{2}$ - (k - 3)t + k = 0

                     ...[taking $x^{2}$ = t]

(i) Now the equation (1) will have four real roots if the equation (2) has both roots as positive and distinct. These values of k are included in the interval (9,∞). Again (1) will have two real roots if either :

a. (2) has equal roots and this equal root is positive

b. (2) has one positive and one negative root.

  Now (2) has has equal roots if k = 1 and k = 9

(ii) k = 1, t = -1 ⇒ no real roots and k = 9, t = 3

  $x^{2}$ = 3 ⇒ x = ± $\sqrt{3}$

Thus, at k = 9 the original equation has exactly two roots.

Again, (2) has one positive and one negative root if k∈ (-∞,0)

(iii) Equation (1) will not have any real roots if equation (2) has either no real roots or has both roots negative. Now (2) has no real root if m∈(1,9) and has both roots negative if k ∈(0,1)

   Thus, required values of k are contained in the interval (0,9)