Find the set of real values of x that satisfies both inequalities x^2 - 3x - 10 < 0 and 3 … — JAMB UTME

Find the set of real values of $x$ that satisfies both inequalities $x^2 - 3x - 10 < 0$ and $3 - 2x \le 1$ .

The set of real values satisfying both inequalities is

1 \le x < 5

Solving the quadratic inequality

x^2 - 3x - 10 < 0

gives the open interval

-2 < x < 5

. Solving the linear inequality

3 - 2x \le 1

gives

-2x \le -2

, which upon dividing by

-2

and reversing the inequality sign becomes

x \ge 1

. Finding the overlapping values that satisfy both inequalities gives

1 \le x < 5

Solve the quadratic inequality

x^2 - 3x - 10 < 0

Factor into

(x - 5)(x + 2) < 0

. Critical values are

x = -2

and

x = 5

. Since the inequality is strictly less than zero, the solution region is

-2 < x < 5

The quadratic expression is negative between its real roots.

Solve the linear inequality

3 - 2x \le 1

Subtract 3 from both sides:

-2x \le -2

. Divide by

-2

and flip the inequality sign:

x \ge 1

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

Find the intersection of the two solution sets.

Combine

-2 < x < 5

and

x \ge 1

to get

1 \le x < 5

Values of

x

must satisfy both conditions simultaneously.

Solving simultaneous linear and quadratic inequalities

Question