You visited us 1 times! Enjoying our articles? Unlock Full Access!

Algebra of Complex Numbers

For what valu...

Question

For what values of $k$ is the inequality $x^{2} - (k - 3) x - k + 6 > 0$ valid for all real $x$ ?

Open in App

Solution

$x^{2} - (k - 3) x - k + 6 > 0, \forall x \in R ∴ D < 0 {(k - 3)}^{2} + 4. (k - 6) < 0 k^{2} - 6 k + 9 + 4 k - 24 < 0 k^{2} - 2 k - 15 < 0 (k - 5) (k + 3) < 0 k \in (- 3, 5)$

Suggest Corrections

Similar questions

k

2 x^{2} - 4 k^{2} x - k^{2} + 1 > 0

x

k

x^{2} - 2 (4 k - 1) x + 15 k^{2} - 2 k - 7 > 0

x

k

x^{2} - 2 (4 k - 1) x + 15 k^{2} - 2 k - 7 > 0

x

^{'} k^{'}

x^{2} - 2 (3 k - 1) x + 8 k^{2} - 7 > 0

x

R

k

∣ ∣ ∣ \frac{x^{2} + k x + 1}{x^{2} + x + 1} ∣ ∣ ∣ < 3

x

Join BYJU'S Learning Program