Linear and Quadratic Inequalities Practice Questions — JAMB UTME

Find the number of integer values of $x$ that satisfy the inequality $3x^2 - 10x - 8 \leq 0$ .

Show answer & explanation

Answer: 5

Answer

The number of integer values of

x

satisfying the inequality is 5.

Factoring

3x^2 - 10x - 8 \leq 0

gives

(3x + 2)(x - 4) \leq 0

. The region where the quadratic expression is non-positive lies between the roots

x = -\frac{2}{3}

and

x = 4

, yielding

-\frac{2}{3} \leq x \leq 4

. The integers falling within this closed interval are

0, 1, 2, 3,

and

4

, giving a total of 5 integer solutions.

Step-by-Step Solution

1

Factor the quadratic expression

(3x + 2)(x - 4) \leq 0

Factoring allows us to find the critical boundary values of the inequality.

2

Find the critical values (roots)

x = -\frac{2}{3}

and

x = 4

Setting each linear factor to zero determines where the expression changes sign.

3

Determine the solution set interval

-\frac{2}{3} \leq x \leq 4

Since the coefficient of

x^2

is positive, the quadratic curve is convex (U-shaped), so the expression is less than or equal to zero between the roots.

4

List and count the integer solutions

Integers:

0, 1, 2, 3, 4

(Total = 5)

The smallest integer greater than or equal to

-\frac{2}{3}

is

0

, and the largest integer less than or equal to

4

is

4

.

Key Concept

Solving quadratic inequalities and identifying integer solutions within a continuous range.

Question 2Question →

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

1 \le x < 5

-2 < x \le 1

x < -2 \text{ or } x > 5

-2 < x < 5

Show answer & explanation

Answer: $1 \le x < 5$

Answer

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

.

Step-by-Step Solution

1

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.

2

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.

3

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.

Key Concept

Solving simultaneous linear and quadratic inequalities

Question 3Question →

Which of the following represents the complete set of real values of $x$ that satisfy the inequality $\frac{x^2 - x - 6}{x - 1} \le 0$ ?

x \le -2

or

1 < x \le 3

x \le -2

or

1 \le x \le 3

-2 \le x < 1

or

x \ge 3

-2 \le x \le 3

with

x \neq 1

Show answer & explanation

Answer: $x \le -2$ or $1 < x \le 3$

Answer

The complete solution set is

x \le -2

or

1 < x \le 3

.

The expression

\frac{(x-3)(x+2)}{x-1}

evaluates to a non-positive value (

\le 0

) when the numerator and denominator have opposite signs or when the numerator is zero. Evaluating across the critical boundaries

x = -2, 1, 3

while excluding

x = 1

yields the solution set

x \le -2

or

1 < x \le 3

.

Step-by-Step Solution

1

Factor the quadratic numerator and state the domain restriction.

The numerator factors as

x^2 - x - 6 = (x - 3)(x + 2)

, giving the rational inequality

\frac{(x - 3)(x + 2)}{x - 1} \le 0

with

x \neq 1

.

Factoring isolates the critical boundary points where the expression can change sign.

2

Identify all critical numbers.

The critical values are

x = -2

,

x = 1

, and

x = 3

.

These points partition the real number line into sub-intervals.

3

Test points in each interval to determine the sign of the rational function

f(x) = \frac{(x - 3)(x + 2)}{x - 1}

.

For

x < -2

,

f(x) \le 0

; for

-2 < x < 1

,

f(x) > 0

; for

1 < x < 3

,

f(x) \le 0

; for

x > 3

,

f(x) > 0

.

Determining where the function is negative or zero identifies the regions satisfying

\le 0

.

4

Combine intervals and include non-undefined endpoints.

Endpoints

x = -2

and

x = 3

make the numerator zero (included), while

x = 1

makes the denominator zero (excluded). Thus,

x \le -2

or

1 < x \le 3

.

Division by zero must be excluded from the solution set.

Key Concept

Solving rational inequalities by finding critical values, using test intervals, and respecting domain restrictions.

Question 4Question →

What is the range of values of $x$ that satisfies the inequality $3 - 2x > 9$ ?

x < -3

x > -3

x < 3

x > 3

Show answer & explanation

Answer: $x < -3$

Answer

x < -3

Subtracting 3 from both sides yields

-2x > 6

. Dividing both sides by

-2

and reversing the inequality symbol gives

x < -3

.

Step-by-Step Solution

1

Subtract 3 from both sides of the inequality.

-2x > 6

To isolate the variable term

-2x

on the left-hand side.

2

Divide both sides by

-2

and reverse the inequality sign.

x < -3

Dividing an inequality by a negative number requires reversing the direction of the inequality sign.

Key Concept

Reversing inequality signs upon multiplication or division by negative numbers

Estimated Time:45s

Question 5Question →

Find the set of real values of $x$ that satisfies the inequality $\frac{3 - x}{x + 2} \geq 1$ .

-2 < x \leq \frac{1}{2}

-2 \leq x \leq \frac{1}{2}

x < -2 \text{ or } x \geq \frac{1}{2}

x \geq \frac{1}{2}

Show answer & explanation

Answer: $-2 < x \leq \frac{1}{2}$

Answer

-2 < x \leq \frac{1}{2}

Subtracting 1 from both sides yields

\frac{1 - 2x}{x + 2} \geq 0

. The critical points are

x = \frac{1}{2}

and

x = -2

. Testing values shows that the fraction is positive for

-2 < x < \frac{1}{2}

and equal to zero at

x = \frac{1}{2}

. Since

x = -2

causes division by zero, it is excluded from the interval, giving

-2 < x \leq \frac{1}{2}

.

Step-by-Step Solution

1

Subtract 1 from both sides of the inequality to set one side to zero.

\frac{3 - x}{x + 2} - 1 \geq 0

Direct cross-multiplication is invalid because the sign of

(x + 2)

depends on

x

.

2

Combine the terms over a common denominator.

\frac{(3 - x) - (x + 2)}{x + 2} \geq 0 \implies \frac{1 - 2x}{x + 2} \geq 0

Simplifying the numerator yields a clear rational inequality expression.

3

Identify the critical points and domain restrictions.

Numerator critical point:

x = \frac{1}{2}

; Denominator restriction:

x \neq -2

.

The quotient changes sign around

x = \frac{1}{2}

and

x = -2

, and division by zero is undefined.

4

Test the intervals

(-\infty, -2)

,

(-2, \frac{1}{2}]

, and

(\frac{1}{2}, \infty)

.

For

x \in (-2, \frac{1}{2}]

, the expression

\frac{1 - 2x}{x + 2}

is non-negative.

When

x = 0

,

\frac{1}{2} > 0

(positive). Outside this interval, the ratio is negative.

Key Concept

Solving Rational and Linear/Quadratic Inequalities

Estimated Time:2m 0s

Question 6Question →

Find the number of integers that satisfy the compound inequality $-3 < 2x + 1 \le 9$ .

Show answer & explanation

Answer: 6

Answer

The number of integer solutions satisfying the inequality is 6.

Subtracting 1 across

-3 < 2x + 1 \le 9

gives

-4 < 2x \le 8

. Dividing by 2 yields

-2 < x \le 4

. The integers in this interval are

-1, 0, 1, 2, 3,

and

4

, giving a total of 6 integer solutions.

Step-by-Step Solution

1

Subtract 1 from all parts of the compound inequality.

-4 < 2x \le 8

Isolate the variable term

2x

in the middle.

2

Divide all parts by 2.

-2 < x \le 4

Solve for

x

by undoing the coefficient of 2.

3

Identify the set of integer solutions within the interval

(-2, 4]

.

x \in \{-1, 0, 1, 2, 3, 4\}

The endpoint

-2

is excluded due to the strict inequality (

<

), while the endpoint

4

is included due to the inclusive inequality (

\\le

).

4

Count the elements in the solution set.

6

There are 6 distinct integer values in the set.

Key Concept

Solving compound linear inequalities and identifying integer solution sets.

Question 7Question →

Which of the following ranges of $x$ satisfies the quadratic inequality $2x^2 - 7x + 3 < 0$ ?

\frac{1}{2} < x < 3

x < \frac{1}{2} \text{ or } x > 3

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

x < -3 \text{ or } x > -\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{2} < x < 3$

Answer

\frac{1}{2} < x < 3

Factorizing

2x^2 - 7x + 3 < 0

gives

(2x - 1)(x - 3) < 0

. The roots of the quadratic equation

(2x - 1)(x - 3) = 0

are

x = \frac{1}{2}

and

x = 3

. Because the coefficient of

x^2

is positive, the quadratic curve opens upwards and is strictly negative between the two roots. Therefore, the inequality is satisfied when

\frac{1}{2} < x < 3

.

Step-by-Step Solution

1

Factorize the quadratic expression

2x^2 - 7x + 3 = (2x - 1)(x - 3)

Factorization allows us to find the critical values of the inequality.

2

Determine the critical values by setting the factored expression to zero

x = \frac{1}{2}

and

x = 3

These points mark the boundary values where the quadratic expression changes sign.

3

Test intervals or apply sign analysis for strict inequality

< 0

The product

(2x - 1)(x - 3)

is negative for

\frac{1}{2} < x < 3

Since the leading coefficient is positive (

2 > 0

), the parabola opens upwards and takes negative values strictly between its real roots.

Key Concept

Solving Quadratic Inequalities by Factorization and Interval Sign Analysis

Estimated Time:1m 15s

Question 8Question →

Find the number of non-negative integer values of $x$ that satisfy the quadratic inequality $x^2 - 5x - 14 \le 0$ .

Show answer & explanation

Answer: 8

Answer

8

Factoring the quadratic expression gives

(x - 7)(x + 2) \le 0

, which evaluates to the solution interval

-2 \le x \le 7

. Restricting this interval to non-negative integers (

x \ge 0

) yields the set

\{0, 1, 2, 3, 4, 5, 6, 7\}

, which contains exactly 8 values.

Step-by-Step Solution

1

Factor the quadratic equation

x^2 - 5x - 14 = 0

(x - 7)(x + 2) = 0

, yielding critical roots at

x = 7

and

x = -2

Finding the roots determines the boundary points for the quadratic inequality.

2

Determine the solution set for the inequality

x^2 - 5x - 14 \le 0

-2 \le x \le 7

The quadratic expression is negative or zero between its two real roots.

3

Identify and count the non-negative integers in the interval

[-2, 7]

The non-negative integers are

0, 1, 2, 3, 4, 5, 6, 7

, giving a total of 8 values.

Non-negative integers consist of zero and all positive whole numbers within the solution range.

Key Concept

Solving quadratic inequalities and identifying discrete non-negative integer solution sets

Question 9Question →

Find the maximum integer value of $m$ for which the quadratic inequality $x^2 - mx + 9 > 0$ holds for all real values of $x$ .

Show answer & explanation

Answer: 5

Answer

The maximum integer value of

m

is

5

.

For the quadratic expression

x^2 - mx + 9

to remain strictly positive for all real values of

x

, the quadratic curve must lie completely above the x-axis. Because the coefficient of

x^2

is positive (

1 > 0

), this requires the discriminant to be strictly negative (

D < 0

). Evaluating

b^2 - 4ac < 0

gives

m^2 - 36 < 0

, which simplifies to

-6 < m < 6

. The largest integer strictly less than

6

is

5

.

Step-by-Step Solution

1

Determine the condition for the quadratic expression to be positive for all real values of

x

.

Since the leading coefficient is

1 > 0

, the condition is that the discriminant

D = b^2 - 4ac < 0

.

A parabola opening upward lies entirely above the horizontal axis when it has no real roots.

2

Calculate the discriminant using the coefficients of the quadratic expression.

D = (-m)^2 - 4(1)(9) = m^2 - 36 < 0

.

Here

a = 1

,

b = -m

, and

c = 9

.

3

Solve the quadratic inequality for

m

.

m^2 - 36 < 0 \implies -6 < m < 6

.

The roots of

m^2 - 36 = 0

are

m = -6

and

m = 6

, and the expression is negative strictly between these boundary values.

4

Determine the maximum integer value within the open interval

(-6, 6)

.

The maximum integer value is

5

.

The boundary value

6

is excluded by the strict inequality

m < 6

.

Key Concept

Quadratic Inequalities and Discriminant Conditions for Positive Definiteness

Question 10Question →

How many integer values of $x$ satisfy both the linear inequality $\frac{3 - 2x}{5} \ge \frac{x + 4}{2}$ and the quadratic inequality $x^2 + 4x - 5 \le 0$ ?

4

3

5

2

Show answer & explanation

Answer: 4

Answer

4

Solving the linear inequality yields

x \le -\frac{14}{9} \approx -1.56

. Solving the quadratic inequality gives

-5 \le x \le 1

. The overlap between both sets is

-5 \le x \le -\frac{14}{9}

. The integers falling within this interval are

-5, -4, -3

, and

-2

, making 4 valid integer solutions in total.

Step-by-Step Solution

1

Solve the linear inequality

\frac{3 - 2x}{5} \ge \frac{x + 4}{2}

.

2(3 - 2x) \ge 5(x + 4) \implies 6 - 4x \ge 5x + 20 \implies -9x \ge 14 \implies x \le -\frac{14}{9} \approx -1.56

.

Clear denominators by multiplying by 10 and reverse the inequality sign when dividing both sides by

-9

.

2

Solve the quadratic inequality

x^2 + 4x - 5 \le 0

.

(x + 5)(x - 1) \le 0 \implies -5 \le x \le 1

.

Factorize the quadratic expression to find critical points at

x = -5

and

x = 1

. The region where the product is non-positive is between the roots.

3

Determine the intersection of the two solution sets.

-5 \le x \le -\frac{14}{9}

.

Combine the conditions

x \le -1.56

and

-5 \le x \le 1

to find the set of values satisfying both inequalities simultaneously.

4

Count the integer values within the intersection set

[-5, -1.56]

.

The integers are

-5, -4, -3, -2

, giving a total of 4 integers.

Identify all whole numbers within the combined solution interval.

Key Concept

Simultaneous Linear and Quadratic Inequalities

Question 11Question →

What is the solution set for the linear inequality $\frac{1 - 4x}{3} \le \frac{x + 7}{2}$ ?

x \ge -\frac{19}{11}

x \le -\frac{19}{11}

x \ge \frac{19}{11}

x \le \frac{23}{5}

Show answer & explanation

Answer: $x \ge -\frac{19}{11}$

Answer

x \ge -\frac{19}{11}

Multiplying the inequality by 6 clears denominators to give

2(1 - 4x) \le 3(x + 7)

. Expanding gives

2 - 8x \le 3x + 21

, which simplifies to

-11x \le 19

. Dividing both sides by

-11

flips the inequality sign, yielding

x \ge -\frac{19}{11}

.

Step-by-Step Solution

1

Clear the denominators by multiplying both sides of the inequality by the lowest common multiple of 3 and 2, which is 6.

6 \cdot \left(\frac{1 - 4x}{3}\right) \le 6 \cdot \left(\frac{x + 7}{2}\right) \implies 2(1 - 4x) \le 3(x + 7)

Eliminating fractions simplifies the algebraic manipulation.

2

Expand both sides by distributing the factors.

2 - 8x \le 3x + 21

Removing parentheses allows grouping of like terms.

3

Rearrange terms by collecting terms containing

x

on the left and constants on the right.

-8x - 3x \le 21 - 2 \implies -11x \le 19

Isolating the variable term prepares for the final division step.

4

Divide both sides by

-11

and reverse the direction of the inequality sign.

x \ge -\frac{19}{11}

Dividing or multiplying an inequality by a negative real number requires flipping the inequality sign.

Key Concept

Solving linear inequalities involving fractions and reversing the inequality sign when dividing by a negative number.

Estimated Time:1m 30s

Question 12Question →

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

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

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

or

x \ge 3

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

x \le -3

or

x \ge \frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2} \le x \le 3$

Answer

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

Factoring

2x^2 - 5x - 3

yields

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

. Setting the factors to zero gives roots

x = -1/2

and

x = 3

. Because the coefficient of

x^2

is positive, the quadratic curve opens upwards and is less than or equal to zero in the closed interval between the roots, resulting in

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

.

Step-by-Step Solution

1

Factor the quadratic expression

2x^2 - 5x - 3

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

Finding the factors helps identify the critical points (roots) of the inequality.

2

Determine the critical points by setting each factor equal to zero

x = -\frac{1}{2}

and

x = 3

The critical points divide the real number line into test intervals.

3

Determine the region where

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

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

Since the quadratic coefficient is positive (

2 > 0

), the parabola opens upwards, so the function values are less than or equal to zero between the two roots.

Key Concept

Solving Quadratic Inequalities by Factorisation

Question 13Question →

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

Show answer & explanation

Answer: 5

Answer

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

or

k > 4

. Intersecting with

k > 2

results in

k > 4

. The smallest integer greater than 4 is 5.

Step-by-Step Solution

1

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

).

2

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.

3

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.

Key Concept

Condition for Positive Definite Quadratic Inequalities

Question 14Question →

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 $x$ ?

k > -1

-2 < k < -1

k < -1

k > -2

Show answer & explanation

Answer: $k > -1$

Answer

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

is

k > -1

.

Step-by-Step Solution

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

).

If

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.

2

Apply the first condition

a > 0

to the coefficient of

x^2

.

k + 2 > 0 \implies k > -2

.

Ensures the parabola opens upwards.

3

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.

4

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.

Key Concept

Conditions for positive definiteness of quadratic expressions

Question 15Question →

How many positive integer values of $x$ satisfy the quadratic inequality $x^2 - 4x - 5 < 0$ ?

Show answer & explanation

Answer: 4

Answer

There are 4 positive integer values of

x

that satisfy the inequality.

Factoring

x^2 - 4x - 5 < 0

yields

(x - 5)(x + 1) < 0

, giving the real solution interval

-1 < x < 5

. Filtering for positive integers (

x \in \mathbb{Z}^+

) yields the set

\{1, 2, 3, 4\}

, which contains exactly 4 values.

Step-by-Step Solution

1

Factor the quadratic inequality

(x - 5)(x + 1) < 0

Factoring helps find the boundary roots of the quadratic equation.

2

Determine the solution set interval

-1 < x < 5

For a quadratic inequality of the form

(x - a)(x - b) < 0

with

a < b

, the solution interval is

a < x < b

.

3

List the positive integer solutions in the range

-1 < x < 5

x \in \{1, 2, 3, 4\}

Positive integers are whole numbers strictly greater than 0.

4

Count the number of positive integers

4

Counting the elements in the set

\{1, 2, 3, 4\}

gives a total of 4.

Key Concept

Solving quadratic inequalities and finding valid integer solutions within a target domain.

Estimated Time:45s

Question 16Question →

Determine the number of integer values of $x$ that satisfy the quadratic inequality $3x^2 - 14x - 5 < 0$ .

Show answer & explanation

Answer: 5

Answer

The total number of integer values satisfying the inequality is 5.

Factoring

3x^2 - 14x - 5 < 0

gives

(3x + 1)(x - 5) < 0

. The critical roots are

x = -\frac{1}{3}

and

x = 5

. Since the parabola opens upward, the expression is negative between the roots, yielding the interval

-\frac{1}{3} < x < 5

. The integer values contained in this interval are

0, 1, 2, 3,

and

4

, which total 5 values.

Step-by-Step Solution

1

Factor the quadratic expression

(3x + 1)(x - 5) < 0

Factoring identifies the critical values where the quadratic expression changes sign.

2

Determine the critical points

x = -\frac{1}{3}

and

x = 5

Setting each factor to zero gives the boundary roots of the equation.

3

Formulate the solution interval

-\frac{1}{3} < x < 5

Since the coefficient of

x^2

is positive, the quadratic curve is below the x-axis strictly between the two roots.

4

List and count the integer solutions

The integers are

0, 1, 2, 3, 4

, yielding 5 integer solutions.

Counting integers strictly greater than

-\frac{1}{3}

and strictly less than

5

.

Key Concept

Solving quadratic inequalities and finding integer solution counts

Estimated Time:1m 30s

Question 17Question →

Find the smallest integer $x$ that satisfies the linear inequality $5 - 3x \le -7$ .

Show answer & explanation

Answer: 4

Answer

The smallest integer value of

x

satisfying the inequality is

4

.

Subtracting 5 from both sides of

5 - 3x \le -7

gives

-3x \le -12

. Dividing both sides by

-3

requires reversing the inequality sign to obtain

x \ge 4

. Therefore, the smallest integer value in the solution set is 4.

Step-by-Step Solution

1

Subtract 5 from both sides of the inequality to isolate the variable term

-3x \le -12

Subtracting a constant from both sides maintains the inequality direction.

2

Divide both sides by

-3

and flip the inequality sign

x \ge 4

Dividing an inequality by a negative number reverses the direction of the inequality symbol.

3

Identify the smallest integer satisfying the condition

4

The solution set contains all real numbers greater than or equal to 4, making 4 the minimum integer value.

Key Concept

Solving Linear Inequalities with Negative Coefficients

Estimated Time:45s

Question 18Question →

Find the sum of all integer values of $x$ that satisfy the system of inequalities $3x + 1 > 7$ and $x^2 - 4x - 21 \le 0$ .

Show answer & explanation

Answer: 25

Answer

25

Solving

3x + 1 > 7

gives

x > 2

. Solving

x^2 - 4x - 21 \le 0

gives

-3 \le x \le 7

. The combined condition is

2 < x \le 7

. The integer values satisfying this condition are 3, 4, 5, 6, and 7. The sum of these integers is

3 + 4 + 5 + 6 + 7 = 25

.

Step-by-Step Solution

1

Solve the linear inequality

3x + 1 > 7

.

x > 2

Subtracting 1 from both sides gives

3x > 6

, and dividing by 3 yields

x > 2

.

2

Solve the quadratic inequality

x^2 - 4x - 21 \le 0

.

-3 \le x \le 7

Factoring the quadratic expression gives

(x - 7)(x + 3) \le 0

. The roots are

-3

and

7

, and the quadratic curve is non-positive between these roots.

3

Determine the intersection of the solution sets

x > 2

and

-3 \le x \le 7

.

2 < x \le 7

The lower bound is determined by the strict linear constraint

x > 2

, while the upper bound is determined by the inclusive quadratic constraint

x \le 7

.

4

List the integer values of

x

within

2 < x \le 7

and compute their sum.

Integers are

3, 4, 5, 6, 7

; their sum is

25

.

The strict inequality excludes

2

, while the inclusive inequality includes

7

.

Key Concept

System of linear and quadratic inequalities

Question 19Question →

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

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

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

or

x \ge 3

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

x \le -3

or

x \ge \frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2} \le x \le 3$

Answer

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

by

-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

.

Step-by-Step Solution

1

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.

2

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)

.

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.

4

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.

Key Concept

Quadratic Inequalities and Sign Reversal

Estimated Time:2m 0s

Question 20Question →

What is the solution set for the linear inequality $4 - 3x \ge 19$ ?

x \le -5

x \ge -5

x \le 5

x \ge 5

Show answer & explanation

Answer: $x \le -5$

Answer

x \le -5

Subtracting

4

from both sides simplifies the inequality to

-3x \ge 15

. Dividing both sides by

-3

requires reversing the direction of the inequality sign, yielding

x \le -5

.

Step-by-Step Solution

1

Subtract 4 from both sides of the inequality

-3x \ge 15

Isolate the variable term

-3x

on the left side of the inequality.

2

Divide both sides by

-3

and reverse the inequality direction

x \le -5

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

\ge

to

\le

.

Key Concept

Reversing inequality signs when dividing by a negative quantity

Estimated Time:45s

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