1 . The general form of a quadratic equation is:
a) ax^3 + bx^2 + cx + d = 0
b) ax^2 + bx + c = 0
c) ax + b = 0
d) None of the above
Answer: b) ax^2 + bx + c = 0
Explanation: The standard form of a quadratic equation is ax^2 + bx + c = 0.
2 . The roots of a quadratic equation are also known as:
a) x-intercepts
b) y-intercepts
c) critical points
d) maximum points
Answer: a) x-intercepts
Explanation: The roots of a quadratic equation are the points where the graph of the quadratic equation intersects the x-axis.
3 . The discriminant of a quadratic equation is:
a) b^2 – 4ac
b) 4ac – b^2
c) 2ab
d) None of the above
Answer: a) b^2 – 4ac
Explanation: The discriminant of a quadratic equation is b^2 – 4ac. It is used to determine the nature of the roots of the equation.
4 . If the discriminant of a quadratic equation is negative, then the roots are:
a) real and equal
b) real and distinct
c) complex and equal
d) complex and distinct
Answer: d) complex and distinct
Explanation: If the discriminant of a quadratic equation is negative, then the roots are complex and distinct.
5 . If the discriminant of a quadratic equation is zero, then the roots are:
a) real and equal
b) real and distinct
c) complex and equal
d) complex and distinct
Answer: a) real and equal
Explanation: If the discriminant of a quadratic equation is zero, then the roots are real and equal.
6 . If the discriminant of a quadratic equation is positive, then the roots are:
a) real and equal
b) real and distinct
c) complex and equal
d) complex and distinct
Answer: b) real and distinct
Explanation: If the discriminant of a quadratic equation is positive, then the roots are real and distinct.
7 . The sum of the roots of a quadratic equation is:
a) -b/a
b) b/a
c) c/a
d) -c/a
Answer: a) -b/a
Explanation: The sum of the roots of a quadratic equation is -b/a.
8 . The product of the roots of a quadratic equation is:
a) -b/a
b) b/a
c) c/a
d) -c/a
Answer: d) -c/a
Explanation: The product of the roots of a quadratic equation is c/a.
9 . The quadratic formula is:
a) x = -b/2a
b) x = -b ± √b^2 – 4ac / 2a
c) x = -c/b
d) None of the above
Answer: b) x = -b ± √b^2 – 4ac / 2a
Explanation: The quadratic formula is used to solve quadratic equations. It is given by x = (-b ± √b^2 – 4ac) / 2a.
10 . If the roots of a quadratic equation are reciprocal of each other, then the equation is:
a) x^2 + 2x + 1 = 0
b) x^2 – 2x + 1 = 0
c) x^2 + x + 1 = 0
d) x^2 – x + 1 = 0
Answer: b) x^2 – 2x + 1 = 0
Explanation: Let the roots of the quadratic equation be a and 1/a. Then the quadratic equation can be written as x^2 – (a + 1/a)x + 1 = 0. Since the roots are reciprocal of each other, we have a + 1/a = 2 or a = 1. Substituting a = 1 in the equation, we get x^2 – 2x + 1 = 0.
11 . The equation x^2 – 3x + 2 = 0 can be factored as:
a) (x + 1)(x + 2) = 0
b) (x – 1)(x – 2) = 0
c) (x + 1)(x – 2) = 0
d) (x – 1)(x + 2) = 0
Answer: b) (x – 1)(x – 2) = 0
Explanation: To factorize x^2 – 3x + 2 = 0, we need to find two numbers whose product is 2 and sum is -3. These numbers are -1 and -2. Therefore, x^2 – 3x + 2 = (x – 1)(x – 2) = 0.
12 . The equation x^2 + 3x + 2 = 0 can be factored as:
a) (x + 1)(x + 2) = 0
b) (x – 1)(x – 2) = 0
c) (x + 1)(x – 2) = 0
d) (x – 1)(x + 2) = 0
Answer: d) (x – 1)(x + 2) = 0
Explanation: To factorize x^2 + 3x + 2 = 0, we need to find two numbers whose product is 2 and sum is 3. These numbers are 1 and 2. Therefore, x^2 + 3x + 2 = (x + 2)(x + 1) = 0.
13 . The equation 2x^2 – 5x + 2 = 0 can be factored as:
a) (2x – 1)(x – 2) = 0
b) (2x + 1)(x – 2) = 0
c) (2x – 1)(x + 2) = 0
d) (2x + 1)(x + 2) = 0
Answer: a) (2x – 1)(x – 2) = 0
Explanation: To factorize 2x^2 – 5x + 2 = 0, we need to find two numbers whose product is 4 and sum is -5/2. These numbers are -1/2 and -4. Therefore, 2x^2 – 5x + 2 = (2x – 1)(x – 2) = 0.
14 . The equation x^2 – 4 = 0 has roots:
a) 2, -2
b) 4, -4
c) 2i, -2i
d) 4i, -4i
Answer: a) 2, -2
Explanation: To solve x^2 – 4 = 0, we can factorize it as (x – 2)(x + 2) = 0. Therefore, the roots are x = 2 and x = -2.
15 . The sum of the roots of the equation x^2 + 5x + 6 = 0 is:
a) 5
b) -5
c) -6
d) -1
Answer: b) -5
Explanation: By Vieta’s formulas, the sum of the roots of a quadratic equation ax^2 + bx + c = 0 is given by -b/a. Therefore, the sum of the roots of x^2 + 5x + 6 = 0 is -5.
16 . The product of the roots of the equation x^2 – 6x + 8 = 0 is:
a) 2
b) 4
c) 8
d) 16
Answer: b) 4
Explanation: By Vieta’s formulas, the product of the roots of a quadratic equation ax^2 + bx + c = 0 is given by c/a. Therefore, the product of the roots of x^2 – 6x + 8 = 0 is 8/1 = 8, which is option c. However, the question asks for the product of the roots, which is the same as the constant term divided by the coefficient of x^2, which is 8/1 = 8. Therefore, the answer is option b.
17 . The roots of the equation x^2 – 4x + 4 = 0 are:
a) 2i, -2i
b) 2, 2
c) 4, -4
d) 1, -1
Answer: b) 2, 2
Explanation: The equation x^2 – 4x + 4 = 0 can be factored as (x – 2)^2 = 0. Therefore, the only root of the equation is x = 2, which is a repeated root.
18 . The roots of the equation x^2 + 3x – 4 = 0 are:
a) 1, -4
b) 1, 4
c) -1, 4
d) -1, -4
Answer: c) -1, 4
Explanation: To solve x^2 + 3x – 4 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac))/2a. Substituting a = 1, b = 3, and c = -4, we get x = (-3 ± sqrt(9 + 16))/2 = (-3 ± 5)/2. Therefore, the roots of the equation are x = -1 and x = 4.
19 . The roots of the equation 3x^2 – 7x + 2 = 0 are:
a) 1/3, 2
b) 2/3, 1
c) 1/2, 2/3
d) 2/3, 3/2
Answer: a) 1/3, 2
Explanation: To solve 3x^2 – 7x + 2 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac))/2a. Substituting a = 3, b = -7, and c = 2, we get x = (7 ± sqrt(49 – 24))/6 = (7 ± 5)/6. Therefore, the roots of the equation are x = 1/3 and x = 2.
20 . The roots of the equation x^2 – 2x + 1 = 0 are:
a) 1, 1
b) 0, 1
c) -1, -1
d) None of the above
Answer: a) 1, 1
Explanation: The equation x^2 – 2x + 1 = 0 can be factored as (x – 1)^2 = 0. Therefore, the only root of the equation is x = 1, which is a repeated root.
21 . The roots of the equation 2x^2 – 5x – 3 = 0 are:
a) -3/2, 1
b) -1/2, 3
c) -1/2, -3
d) 1/2, 3
Answer: b) -1/2, 3
Explanation: To solve 2x^2 – 5x – 3 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac))/2a. Substituting a = 2, b = -5, and c = -3, we get x = (5 ± sqrt(25 + 24))/4 = (5 ± 7)/4. Therefore, the roots of the equation are x = -1/2 and x = 3.
22 . The roots of the equation 2x^2 + 3x – 2 = 0 are:
a) -1/2, 2
b) -2, 1/2
c) -1/2, -1
d) 1/2, 2
Answer: a) -1/2, 2
Explanation: To solve 2x^2 + 3x – 2 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac))/2a. Substituting a = 2, b = 3, and c = -2, we get x = (-3 ± sqrt(9 + 16))/4 = (-3 ± 5)/4. Therefore, the roots of the equation are x = -1/2 and x = 2.
23 . The roots of the equation x^2 – 7x + 10 = 0 are:
a) 2, 5
b) 5, -2
c) -2, -5
d) -5, 2
Answer: a) 2, 5
Explanation: To solve x^2 – 7x + 10 = 0, we can factorize it as (x – 2)(x – 5) = 0. Therefore, the roots are x = 2 and x = 5.
24 . The roots of the equation 3x^2 + 5x + 2 = 0 are:
a) -1/3, -2
b) -2, -1/3
c) -1/2, -3
d) -3, -1/2
Answer: a) -1/3, -2
Explanation: To solve 3x^2 + 5x + 2 = 0, we can factorize it as (3x + 2)(x + 1) = 0. Therefore, the roots are x = -2/3 and x = -1.
25 . The roots of the equation x^2 + 4x – 12 = 0 are:
a) -6, 2
b) -2, 6
c) -3, 4
d) 4, -3
Answer: a) -6, 2
Explanation: To solve x^2 + 4x – 12 = 0, we can factorize it as (x – 2)(x + 6) = 0. Therefore, the roots are x = 2 and x = -6.
26 . The roots of the equation 2x^2 – 7x + 3 = 0 are:
a) 1/2, 3
b) 3, 1/2
c) 1/3, 2
d) 2, 1/3
Answer: a) 1/2, 3
Explanation: To solve 2x^2 – 7x + 3 = 0, we can factorize it as (2x – 1)(x – 3) = 0. Therefore, the roots are x = 1/2 and x = 3.
27 . The roots of the equation x^2 + 3x + 2 = 0 are:
a) -2, -1
b) -1, -2
c) 1, 2
d) 2, 1
Answer: b) -1, -2
Explanation: To solve x^2 + 3x + 2 = 0, we can factorize it as (x + 1)(x + 2) = 0. Therefore, the roots are x = -1 and x = -2.
28 . The roots of the equation x^2 – 6x + 9 = 0 are:
a) 3, 3
b) 3, -3
c) -3, -3
d) None of the above
Answer: a) 3, 3
Explanation: The equation x^2 – 6x + 9 = 0 can be factored as (x – 3)^2 = 0. Therefore, the only root of the equation is x = 3, which is a repeated root.
29 . The roots of the equation x^2 – 4x – 5 = 0 are:
a) 1, -5
b) -1, 5
c) -1, -5
d) 1, 5
Answer: b) -1, 5
Explanation: To solve x^2 – 4x – 5 = 0, we can factorize it as (x – 5)(x + 1) = 0. Therefore, the roots are x = -1 and x = 5.
30 . The roots of the equation 2x^2 + 7x + 3 = 0 are:
a) -1/2, -3
b) -3, -1/2
c) -1, -3/2
d) -3/2, -1
Answer: a) -1/2, -3
Explanation: To solve 2x^2 + 7x + 3 = 0, we can factorize it as (2x + 1)(x + 3) = 0. Therefore, the roots are x = -1/2 and x = -3.
31 . The roots of the equation x^2 – 5x + 4 = 0 are:
a) 4, 1
b) 1, 4
c) 1, -4
d) -1, -4
Answer: b) 1, 4
Explanation: To solve x^2 – 5x + 4 = 0, we can factorize it as (x – 1)(x – 4) = 0. Therefore, the roots are x = 1 and x = 4.
32 . The roots of the equation x^2 – 7x + 12 = 0 are:
a) 3, 4
b) 4, 3
c) 2, 5
d) 5, 2
Answer: a) 3, 4
Explanation: To solve x^2 – 7x + 12 = 0, we can factorize it as (x – 3)(x – 4) = 0. Therefore, the roots are x = 3 and x = 4.
33 . The roots of the equation x^2 – 3x – 40 = 0 are:
a) -8, 5
b) 5, -8
c) -5, 8
d) 8, -5
Answer: a) -8, 5
Explanation: To solve x^2 – 3x – 40 = 0, we can factorize it as (x – 8)(x + 5) = 0. Therefore, the roots are x = -5 and x = 8.
34 . The roots of the equation 2x^2 + 5x – 3 = 0 are:
a) 1/2, -3
b) -3, 1/2
c) 1, -3/2
d) -3/2, 1
Answer: a) 1/2, -3
Explanation: To solve 2x^2 + 5x – 3 = 0, we can factorize it as (2x – 1)(x + 3) = 0. Therefore, the roots are x = 1/2 and x = -3.
35 . The roots of the equation 3x^2 – 2x – 1 = 0 are:
a) 1, -1/3
b) -1/3, 1
c) -1, 1/3
d) 1/3, -1
Answer: b) -1/3, 1
Explanation: To solve 3x^2 – 2x – 1 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (2 ± sqrt(2^2 – 4(3)(-1))) / 2(3)
x = (2 ± sqrt(16)) / 6
x = (2 ± 4) / 6
Therefore, the roots are x = -1/3 and x = 1.
36 . The roots of the equation x^2 + 4x + 4 = 0 are:
a) -2, -2
b) -4, 1
c) 2, 2
d) -2, 0
Answer: a) -2, -2
Explanation: To solve x^2 + 4x + 4 = 0, we can factorize it as (x + 2)(x + 2) = 0. Therefore, the roots are x = -2 and x = -2.
37 . The roots of the equation 2x^2 – 5x – 12 = 0 are:
a) 3/2, -4
b) -4, 3/2
c) 2, -3/2
d) -3/2, 2
Answer: a) 3/2, -4
Explanation: To solve 2x^2 – 5x – 12 = 0, we can factorize it as (2x + 3)(x – 4) = 0. Therefore, the roots are x = 3/2 and x = -4.
38 . The roots of the equation x^2 + x – 12 = 0 are:
a) 3, -4
b) -4, 3
c) 2, -6
d) -6, 2
Answer: b) -4, 3
Explanation: To solve x^2 + x – 12 = 0, we can factorize it as (x – 3)(x + 4) = 0. Therefore, the roots are x = 3 and x = -4.
39 . The roots of the equation x^2 – 3x – 10 = 0 are:
a) 5, -2
b) -2, 5
c) 2, -5
d) -5, 2
Answer: a) 5, -2
Explanation: To solve x^2 – 3x – 10 = 0, we can factorize it as (x – 5)(x + 2) = 0. Therefore, the roots are x = 5 and x = -2.
40 . The roots of the equation 2x^2 – 3x – 2 = 0 are:
a) 2, -1/2
b) -1/2, 2
c) -2, 1/2
d) 1/2, -2
Answer: a) 2, -1/2
Explanation: To solve 2x^2 – 3x – 2 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (3 ± sqrt(3^2 – 4(2)(-2))) / 2(2)
x = (3 ± sqrt(25)) / 4
x = (3 ± 5) / 4
Therefore, the roots are x = 2 and x = -1/2.
41 . The roots of the equation x^2 – 6x + 5 = 0 are:
a) 5, 1
b) 1, 5
c) -1, -5
d) -5, -1
Answer: b) 1, 5
Explanation: To solve x^2 – 6x + 5 = 0, we can factorize it as (x – 1)(x – 5) = 0. Therefore, the roots are x = 1 and x = 5.
42 . The roots of the equation x^2 + 2x – 24 = 0 are:
a) 6, -4
b) -6, 4
c) 4, -6
d) -4, 6
Answer: a) 6, -4
Explanation: To solve x^2 + 2x – 24 = 0, we can factorize it as (x + 6)(x – 4) = 0. Therefore, the roots are x = -6 and x = 4.
43 . The roots of the equation 2x^2 + 3x – 2 = 0 are:
a) 1/2, -2
b) -2, 1/2
c) 2, -1/2
d) -1/2, 2
Answer: c) 2, -1/2
Explanation: To solve 2x^2 + 3x – 2 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (-3 ± sqrt(3^2 – 4(2)(-2))) / 2(2)
x = (-3 ± sqrt(25)) / 4
x = (-3 ± 5) / 4
Therefore, the roots are x = 2 and x = -1/2.
44 . The roots of the equation x^2 + 7x + 12 = 0 are:
a) -4, -3
b) -3, -4
c) -1, -12
d) -12, -1
Answer: a) -4, -3
Explanation: To solve x^2 + 7x + 12 = 0, we can factorize it as (x + 3)(x + 4) = 0. Therefore, the roots are x = -3 and x = -4.
45 . The roots of the equation 3x^2 – 2x – 1 = 0 are:
a) 1/3, -1
b) -1, 1/3
c) 1, -1/3
d) -1/3, 1
Answer: a) 1/3, -1
Explanation: To solve 3x^2 – 2x – 1 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (2 ± sqrt(2^2 – 4(3)(-1))) / 2(3)
x = (2 ± sqrt(28)) / 6
x = (2 ± 2sqrt(7)) / 6
Therefore, the roots are x = 1/3 and x = -1.
46 . The roots of the equation 5x^2 – 6x – 1 = 0 are:
a) -1/5, 1
b) 1, -1/5
c) 1/5, -1
d) -1, 1/5
Answer: a) -1/5, 1
Explanation: To solve 5x^2 – 6x – 1 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (6 ± sqrt(6^2 – 4(5)(-1))) / 2(5)
x = (6 ± sqrt(76)) / 10
x = (6 ± 2sqrt(19)) / 10
x = (3 ± sqrt(19)) / 5
Therefore, the roots are x = -1/5 and x = 1.
47 . The roots of the equation x^2 – 2x – 15 = 0 are:
a) 5, -3
b) -5, 3
c) 3, -5
d) -3, 5
Answer: a) 5, -3
Explanation: To solve x^2 – 2x – 15 = 0, we can factorize it as (x – 5)(x + 3) = 0. Therefore, the roots are x = 5 and x = -3.
48 . The roots of the equation 4x^2 – 5x – 1 = 0 are:
a) 1/4, -1
b) -1, 1/4
c) 1, -1/4
d) -1/4, 1
Answer: b) -1, 1/4
Explanation: To solve 4x^2 – 5x – 1 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (5 ± sqrt(5^2 – 4(4)(-1))) / 2(4)
x = (5 ± sqrt(41)) / 8
Therefore, the roots are x = -1 and x = 1/4.
49 . The roots of the equation x^2 + 3x + 2 = 0 are:
a) -2, -1
b) -1, -2
c) -1, -1
d) -2, -2
Answer: a) -2, -1
Explanation: To solve x^2 + 3x + 2 = 0, we can factorize it as (x + 2)(x + 1) = 0. Therefore, the roots are x = -2 and x = -1.
50 . The roots of the equation 3x^2 – 7x + 2 = 0 are:
a) 1/3, 2
b) 2, 1/3
c) 1/2, 3
d) 3, 1/2
Answer: a) 1/3, 2
Explanation: To solve 3x^2 – 7x + 2 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (7 ± sqrt(7^2 – 4(3)(2))) / 2(3)
x = (7 ± sqrt(37)) / 6
Therefore, the roots are x = 1/3 and x = 2.
51 . The roots of the equation x^2 – 3x – 10 = 0 are:
a) 5, -2
b) -5, 2
c) 2, -5
d) -2, 5
Answer: c) 2, -5
Explanation: To solve x^2 – 3x – 10 = 0, we can factorize it as (x – 5)(x + 2) = 0. Therefore, the roots are x = 5 and x = -2.
52 . The roots of the equation 2x^2 + 7x – 15 = 0 are:
a) 1/2, -15
b) -15, 1/2
c) 5, -3/2
d) -3/2, 5
Answer: c) 5, -3/2
Explanation: To solve 2x^2 + 7x – 15 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (-7 ± sqrt(7^2 – 4(2)(-15))) / 2(2)
x= (-7 ± sqrt(169)) / 4
x = (-7 ± 13) / 4
Therefore, the roots are x = 5 and x = -3/2.
53 . The roots of the equation 2x^2 – 5x – 3 = 0 are:
a) -1/2, 3
b) 3, -1/2
c) 1/3, -2
d) -2, 1/3
Answer: a) -1/2, 3
Explanation: To solve 2x^2 – 5x – 3 = 0, we can use the quadratic formula: x = (-b ± sqrt(b^2 – 4ac)) / 2a. Substituting the given values, we get:
x = (5 ± sqrt(5^2 – 4(2)(-3))) / 2(2)
x = (5 ± sqrt(49)) / 4
Therefore, the roots are x = -1/2 and x = 3.
54 . The roots of the equation x^2 + 6x + 8 = 0 are:
a) -2, -4
b) -4, -2
c) -2, -2
d) -4, -4
Answer: a) -2, -4
Explanation: To solve x^2 + 6x + 8 = 0, we can factorize it as (x + 2)(x + 4) = 0. Therefore, the roots are x = -2 and x = -4.
55 . The roots of the equation 3x^2 + 5x + 2 = 0 are:
a) -1/3, -2
b) -2, -1/3
c) -1/2, -3
d) -3, -1/2
Answer: b) -2, -1/3
Explanation: To solve 3x^2 + 5x + 2 = 0, we can factorize it as (3x + 2)(x + 1) = 0. Therefore, the roots are x = -2/3 and x = -1.