If the equation $x^{4} + k x^{2} + k = 0$ has exactly two distinct real roots, then the smallest integral value of $| k |$ is

Open in App

Solution

$x^{4} + k x^{2} + k = 0 \dots (1)$
Let $x^{2} = t \Rightarrow t \in [0, \infty)$ ,
$t^{2} + k t + k = 0 \dots (2)$
Let the roots of equation $(2)$ are $t_{1}, t_{2} (t_{1} \leq t_{2})$
Equation $(1)$ will have exactly $2$ distinct real roots iff
(A) $t_{1} < 0, t_{2} > 0$ and
(B) $t_{1} = t_{2} > 0$

Case(A) $: t_{1} < 0, t_{2} > 0$
$t_{1} < 0 < t_{2}$ , so $0$ lies in between the roots,
$f (0) < 0 \Rightarrow k < 0 \dots (3)$

Case (B) $2 : t_{1} = t_{2} > 0$
$(i) D = 0 \Rightarrow k^{2} - 4 k = 0 \Rightarrow k = 0, 4 (i i) - \frac{b}{2 a} > 0 \Rightarrow - \frac{k}{2} > 0 \Rightarrow k < 0 ∴ k \in ϕ \dots (4)$

From equation $(3)$ and $(4)$ ,
$k < 0$

Hence, the minimum value of $| k |$ is $1.$

Suggest Corrections

Similar questions

x^{4} + k x^{2} + k = 0

| k |

(k - 2) x^{2} + 8 x + k + 4 = 0

| l n | x | | = | k - 1 | - 4

x \in R - {0}

x^{4} - (k - 1) x^{2} + (2 - k) = 0

k

| l n | x | | = | k - 1 | - 4

x \in R - {0}

Join BYJU'S Learning Program