For what set of real values of k does the quadratic inequality (k + 2)x^2 - 2(k - 1)x + (k… — JAMB UTME

For what set of real values of $k$ does the quadratic inequality $(k + 2)x^2 - 2(k - 1)x + (k + 5) > 0$ hold for all real values of $x$ ?

The condition holds for

k > -1

A quadratic function is strictly positive for all real values of

x

if and only if its parabola opens upwards (

a > 0

) and it has no real x-intercepts (

\Delta < 0

). Here

a = k + 2 > 0

gives

k > -2

. The discriminant

\Delta = [-2(k-1)]^2 - 4(k+2)(k+5) = -36(k+1)

. Setting

-36(k+1) < 0

requires dividing by

-36

and reversing the inequality sign, giving

k + 1 > 0 \implies k > -1

. The intersection of

k > -2

and

k > -1

k > -1

Identify the conditions required for a quadratic expression

ax^2 + bx + c

to be strictly positive for all real

x

The coefficient of

x^2

must be positive (

a > 0

) and the discriminant must be strictly negative (

\Delta < 0

a < 0

, the parabola opens downwards and yields negative values. If

\Delta \ge 0

, the quadratic has real roots or a double root, taking non-positive values.

Apply the first condition

a > 0

to the coefficient of

x^2

k + 2 > 0 \implies k > -2

Ensures the parabola opens upwards.

Calculate the discriminant

\Delta = b^2 - 4ac

and set

\Delta < 0

\Delta = [-2(k - 1)]^2 - 4(k + 2)(k + 5) = 4(k^2 - 2k + 1) - 4(k^2 + 7k + 10) = 4(-9k - 9) = -36(k + 1)

. Setting

-36(k + 1) < 0

Ensures the quadratic equation has no real roots and does not cross or touch the x-axis.

Solve the linear inequality

-36(k + 1) < 0

and combine with the first condition.

Dividing by

-36

reverses the inequality:

k + 1 > 0 \implies k > -1

. Taking the intersection of

k > -2

and

k > -1

yields

k > -1

Dividing an inequality by a negative number flips the inequality symbol.

Conditions for positive definiteness of quadratic expressions

Question