Quadratic Equation Test

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Quadratic Equation Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

If the roots of the quadratic equation ax² + bx + c = 0 are real and distinct, then _____.

A
b² - 2ab = 0

B
b² - 4ac < 0

C
b² - 4ac > 0

D
b² + 4ac < 0

Solution

ax² + bx + c = 0

b² - 4ac > 0
Question 2
1 / -0

What will be the discriminant of the given quadratic equation?

5x² + x + 7 = 0

A
- 138

B
- 139

C
- 140

D
- 141

Solution

5x² + x + 7 = 0

D = 1 - 4 x 5 x 7

= - 139
Question 3
1 / -0

What will be the discriminant of the given quadratic equation?

2x² + 7x + 5 = 0

A
-9

B
-7

C
7

D
9

Solution

2x² + 7x + 5 = 0

D = 49 - 4 x 2 x 5
= 9
Question 4
1 / -0

If the roots of the quadratic equation ax² + bx + c = 0 are real and equal, then ______.

A
b² – 4ac = 0

B
b² – 4ac + ab > 0

C
b² – 4ac + 2a = 0

D
b² – 4ac < 0

Solution

ax² + bx + c = 0 for equal and real D = 0

D = b² - 4ac = 0
Question 5
1 / -0

Which of the following equations can be obtained after completing the square of x²+ 6x + 3 = 0?

A
(x + 1)² = 5

B
(x + 3)² = 6

C
(x - 2)² = 9

D
(x - 4)² = 7

Solution

x² + 6x + 3 = 0

(x + 3)² + 3 - 9 = 0

(x + 3)² = 6
Question 6
1 / -0

What will be the discriminant of the given quadratic equation?

ax² + bx + c = 0

A
b² – 4a²c

B
b² + 2ab

C
b² – 4ac

D
b² + 4a²

Solution

ax² + bx + c = 0

b² – 4ac
Question 7
1 / -0

Which of the following equations has no real roots?

A
x² + x + 12 = 0

B
x² - x - 1 = 0

C
2x² + 7x + 5 = 0

D
x² + x - 1 = 0

Solution

x² + x + 12 = 0

D = 1 - 4 × (12)

= - 47
Question 8
1 / -0

Which of the following equations can be obtained after completing the square of x² + 4x + 2 = 0?

A
(x + 3)² = 6

B
(x - 2)² = 4

C
(x - 3)² = 2

D
(x + 2)² = 2

Solution

x² + 4x + 2 = 0

(x + 2)² + 2 - 4 = 0

(x + 2)² = 2

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Re-Attempt Weekly Quiz Competition

Quadratic Equation Test - 2

Result

Quadratic Equation Test - 2

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SHARING IS CARING

Question 1

b2 - 2ab = 0

b2 - 4ac < 0

b2 - 4ac > 0

b2 + 4ac < 0

Solution

Question 2

- 138

- 139

- 140

- 141

Solution

Question 3

-9

-7

7

9

Solution

Question 4

b2 – 4ac = 0

b2 – 4ac + ab > 0

b2 – 4ac + 2a = 0

b2 – 4ac < 0

Solution

Question 5

(x + 1)2 = 5

(x + 3)2 = 6

(x - 2)2 = 9

(x - 4)2 = 7

Solution

Question 6

b2 – 4a2c

b2 + 2ab

b2 – 4ac

b2 + 4a2

Solution

Question 7

x2 + x + 12 = 0

x2 - x - 1 = 0

2x2 + 7x + 5 = 0

x2 + x - 1 = 0

Solution

Question 8

(x + 3)2 = 6

(x - 2)2 = 4

(x - 3)2 = 2

(x + 2)2 = 2

Solution

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b² - 2ab = 0

b² - 4ac < 0

b² - 4ac > 0

b² + 4ac < 0

b² – 4ac = 0

b² – 4ac + ab > 0

b² – 4ac + 2a = 0

b² – 4ac < 0

(x + 1)² = 5

(x + 3)² = 6

(x - 2)² = 9

(x - 4)² = 7

b² – 4a²c

b² + 2ab

b² – 4ac

b² + 4a²

x² + x + 12 = 0

x² - x - 1 = 0

2x² + 7x + 5 = 0

x² + x - 1 = 0

(x + 3)² = 6

(x - 2)² = 4

(x - 3)² = 2

(x + 2)² = 2