Find the smallest positive integer k for which the inequality (k - 2)x^2 + 8x + k + 4 > 0 … — JAMB UTME

Find the smallest positive integer $k$ for which the inequality $(k - 2)x^2 + 8x + k + 4 > 0$ holds for all real values of $x$ .

Cevap: 5

For the quadratic expression

(k - 2)x^2 + 8x + k + 4

to be positive for all real values of

x

, two conditions must be satisfied simultaneously: the leading coefficient must be positive (

k - 2 > 0 \implies k > 2

) and the discriminant must be strictly negative (

\Delta < 0

). Calculating the discriminant gives

\Delta = 8^2 - 4(k - 2)(k + 4) = 96 - 8k - 4k^2

. Setting

96 - 8k - 4k^2 < 0

and dividing by

-4

(reversing the inequality) yields

k^2 + 2k - 24 > 0

, which factors as

(k + 6)(k - 4) > 0

. This gives

k < -6

k > 4

. Intersecting with

k > 2

results in

k > 4

. The smallest integer greater than 4 is 5.

Determine the conditions for positivity for all real numbers

k - 2 > 0

and

\Delta < 0

A quadratic

Ax^2 + Bx + C

remains above the x-axis for all real

x

if and only if its parabola opens upwards (

A > 0

) and has no real roots (

\Delta < 0

Set up and solve the discriminant inequality

k^2 + 2k - 24 > 0 \implies (k + 6)(k - 4) > 0

Expanding

8^2 - 4(k - 2)(k + 4) < 0

gives

64 - 4(k^2 + 2k - 8) < 0

, which simplifies to

k^2 + 2k - 24 > 0

after dividing by

-4

and reversing the inequality sign.

Intersect solution sets and find the smallest integer

k = 5

The intersection of

k > 2

and

(k < -6 \text{ or } k > 4)

gives

k > 4

. The smallest integer strictly greater than 4 is 5.

Condition for Positive Definite Quadratic Inequalities

Soru