What is the set of real values of x that satisfies the quadratic inequality -2x^2 + 5x + 3… — JAMB UTME

What is the set of real values of $x$ that satisfies the quadratic inequality $-2x^2 + 5x + 3 \ge 0$ ?

The set of real values of

x

satisfying the inequality is

-\frac{1}{2} \le x \le 3

Multiplying

-2x^2 + 5x + 3 \ge 0

-1

gives

2x^2 - 5x - 3 \le 0

. Factorizing yields

(2x + 1)(x - 3) \le 0

. The product is non-positive between the roots

x = -\frac{1}{2}

and

x = 3

, giving

-\frac{1}{2} \le x \le 3

Multiply or divide the inequality by

-1

to make the leading coefficient positive.

2x^2 - 5x - 3 \le 0

Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.

Factorize the quadratic expression

2x^2 - 5x - 3

(2x + 1)(x - 3) \le 0

Splitting the middle term

-5x

into

-6x + x

gives

2x(x - 3) + 1(x - 3) = (2x + 1)(x - 3)

Find the critical points by setting the expression equal to zero.

x = -\frac{1}{2}

and

x = 3

Critical points mark the boundaries where the sign of the quadratic expression changes.

Determine the interval satisfying the inequality

(2x + 1)(x - 3) \le 0

-\frac{1}{2} \le x \le 3

A quadratic expression with a positive coefficient of

x^2

is less than or equal to zero between its roots.

Quadratic Inequalities and Sign Reversal

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