Free Linear and Quadratic Equations and Inequalities MCQs with Answers

• A. 10
• B. 11
• C. 12
• D. 13

Explanation: Subtracting 2x from both sides gives x - 7 = 5. Adding 7 gives x = 12, so the correct option is 12.

• A. x < -3
• B. x > -3
• C. x > 3
• D. x < 3

Explanation: Subtracting 5 gives -2x > 6. Dividing by the negative number -2 reverses the inequality, giving x < -3.

• A. 2
• B. 3
• C. 4
• D. 5

Explanation: Adding the two equations gives 3x = 9, so x = 3. Substituting in x - y = 2 gives y = 1, and therefore x + y = 4.

• A. 1 and 6
• B. -1 and -6
• C. -2 and -3
• D. 2 and 3

Explanation: The quadratic factors as (x - 2)(x - 3) = 0. Hence, its roots are x = 2 and x = 3.

• A. No real roots
• B. One real root
• C. Two equal real roots
• D. Two distinct real roots

Explanation: The discriminant is b² - 4ac = 4² - 4(2)(5) = -24. Since the discriminant is negative, the equation has no real roots.

• A. 1
• B. 2
• C. 3
• D. 4

Explanation: Completing the square gives x² - 6x + 11 = (x - 3)² + 2. The square is at least zero, so the minimum value is 2 when x = 3.

• A. x ≤ -1 or x ≥ 2
• B. x < -1 or x > 2
• C. -1 ≤ x ≤ 2
• D. -2 ≤ x ≤ 1

Explanation: The critical values are -1 and 2. The product is non-positive between the roots, including the roots, so the solution is -1 ≤ x ≤ 2.

• A. 3
• B. 6
• C. 9
• D. 12

Explanation: Equal roots require the discriminant to be zero: k² - 4(1)(9) = 0. Thus k² = 36, and because k is positive, k = 6.

• A. 7/2
• B. 3
• C. 5/2
• D. 4

Explanation: Multiplying by x - 2 gives x + 1 = 3x - 6. Therefore, 2x = 7 and x = 7/2, which satisfies the restriction x ≠ 2.

• A. -1 < x < 5
• B. x ≤ -1 or x ≥ 5
• C. x < -1 or x > 5
• D. -5 < x < 1

Explanation: The expression factors as (x - 5)(x + 1). Since the parabola opens upward, it is positive outside the roots, giving x < -1 or x > 5.

• A. x = 5
• B. x = 17/5
• C. x = 23/5
• D. x = 11/5

Explanation: Expanding gives 8x - 12 = 3x + 11, so 5x = 23 and x = 23/5. The value 17/5 results from an incorrect simplification of the constant terms.

• A. 7/5
• B. 12/5
• C. 11/5
• D. 9/5

Explanation: From x - y = 1, we get x = y + 1. Substitution gives 3(y + 1) + 2y = 12, so 5y = 9 and y = 9/5.

• A. 1/2 and 3
• B. 1 and 3
• C. −1/2 and 3
• D. 1/2 and −3

Explanation: The equation factors as (2x - 1)(x - 3) = 0. Therefore, x = 1/2 or x = 3; the negative options arise from sign errors during factoring.

• A. p < −8 or p > 8
• B. −8 < p < 8
• C. −8 ≤ p ≤ 8
• D. −4 < p < 4

Explanation: A quadratic has no real roots when its discriminant is negative. Here, p² - 4(2)(8) < 0 gives p² < 64, hence −8 < p < 8.

• A. x = −8
• B. x = 4
• C. x = −4
• D. x = 8

Explanation: For y = ax² + bx + c, the axis of symmetry is x = −b/(2a). Thus x = −8/2 = −4, while x = 4 results from missing the negative sign.

• A. x ≤ −11
• B. x ≥ −11
• C. x < 11
• D. x > 11

Explanation: Subtracting 2x and 4 from both sides gives x ≤ −11. The inequality sign does not reverse because the coefficient of x is positive when the…

• A. −2 < x < 3
• B. x < −2 or x > 3
• C. x ≤ −2 or x ≥ 3
• D. −3 < x < 2

Explanation: The critical values are −2 and 3. The product is positive outside these roots, so the solution is x < −2 or x > 3, with the endpoints…

• A. 37/4
• B. 45/4
• C. 53/4
• D. 81/4

Explanation: By the sum and product of roots, α + β = 9/2 and αβ = 7/2. Therefore, α² + β² = (9/2)² − 2(7/2) = 53/4.

• A. x = 2 or x = −3
• B. x = 3 or x = 2
• C. x = 5 or x = −1
• D. x = 3 or x = −2

Explanation: The absolute value equation gives 2x − 1 = 5 or 2x − 1 = −5. These produce x = 3 and x = −2 respectively.

• A. x < 0 or x > 1
• B. 0 < x < 1
• C. x ≤ 0 or x ≥ 1
• D. x > 0 and x ≠ 1

Explanation: The critical points are x = 0, where the numerator is zero, and x = 1, where the expression is undefined.